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BuGineenng  Science  Series 


DESCRIPTIVE   GEOMETRY 


ENGINEERING   SCIENCE   SERIES 

EDITED  BY 

DUGALD   C.   JACKSON,   C.E. 

Professor  of  Electrical  Engineering 

Massachusetts  Institute  of  TEcnNOLOQY 

Fellow  and  Past  President  A.I.E.E. 

EARLE   R.   HEDRICK,   Ph.D. 

Professok  of  Mathematics,  Univeksitv  of  Missouri 
Member  A.S.M.E. 


DESCRIPTIVE   GEOMETRY 


BY 

GEORGE   YOUNG,   Jr. 

PROFESSOR    OF  ARCHITECTURE 
CORNELL    UNIVERSITY 

AND 

HUBERT   EUGENE   BAXTER 

ASSISTANT    PROFESSOR    OF    ARCHITECTURE 
CORNELL    UNI^^ERSITY 


^  •  /»\  L.  Tr  '• . •  i^*"  o  '* '  * » ' 


THE   MACMILLAN   COMPANY 
1921 

All  rights  reserved 


(^A^i>l 


PRTNTEP    IN    THE    FNITI'D    RTATEB    OF    AMERICA 


Engineering 
Library' 


Copyright,  1921, 

bt  the  macmillan  company. 


Set  up  and  electrotypeA        Published  October,  192 1. 


bA^ 


Kortoooti  5Prf33 

J.  S.  Gushing  Co.  —  Berwir-k  &  Smith  Co. 

Norwood,  Mass.,  U.S.A. 


PREFACE 

In  adding  to  the  already  long  and  excellent  list  of  texts  on 
Descriptive  Geometry,  the  ^^Titers  have  not  attempted  to 
present  any  new  abstract  material  nor  even  to  include  all  of  the 
standard  problems.  On  the  other  hand,  they  have  sought  to 
avoid  a  presentation  which  dwells  solely  on  the  practical  side 
of  the  subject. 

The  value  of  Descriptive  Geometry  in  developing  the  imagi- 
nation, through  the  ability  to  visualize,  has,  for  a  long  time,  been 
fully  recognized,  not  only  in  the  schools,  but  also  by  men  who 
have  won  recognition  and  maturity  in  the  practice  of  their 
professions.  The  average  technical  student,  however,  while 
eager  in  the  pursuit  of  any  subject  which  seems  to  offer  an 
immediate  and  definite  return,  is  reluctant  to  apply  himself  to 
a  subject  whose  value  is  not  at  once  apparent. 

In  preparing  this  text,  the  attempt  has  been  made  to  hold 
the  student's  attention  by  means  of  introductory  paragraphs 
and  other  explanatory  matter  intended  to  show  the  relation  of 
the  principles  under  discussion  to  structural  work.  At  the 
same  time  the  treatment  of  the  various  subjects  is  kept  purely 
abstract  in  order  to  avoid,  in  so  far  as  is  possible,  the  tendency 
of  the  whole  subject  to  degenerate  into  practical  rules  and 
formulas.  Finally,  there  has  been  added  a  set  of  exercises 
designed  to  show  the  application  of  the  abstract  ideas  to 
concrete,  work-a-day  problems.  These  applications  have  been 
grouped  apart  from  the  abstract  problems  in  order  to  make  it 
possible  to  use  them  with  more  or  less  freedom  as  the  case  may 
require,  and  also  to  emphasize  the  secondary  and  dependent 
character  of  such  problems. 

V 

451780 


vi  PREFACE 

Believing  that  the  chief  value  of  Descriptive  Geometry  lies 
in  its  imaginative  quality,  the  ^\Tite^s  have  attempted  so  to 
present  the  subject  matter  as  to  encourage  intuitive  rather  than 
rigidly  formal  methods.  For  this  reason  the  usual  terminology 
has  occasionally  been  avoided.  The  use  of  first  quadrant  solu- 
tions has  been  adopted  as  being  more  natural,  easier  to  visualize, 
and  in  no  wa}^  incompatible  with  actual  practice. 

Some  parts  of  the  book,  notably  the  chapters  on  Curves, 
Topography,  and  Other  Methods  of  Projection,  have  been  intro- 
duced mainly  as  reference  matter,  though  they  might  be  in- 
cluded as  a  part  of  the  regular  course  where  time  permits. 

More  than  the  usual  amount  of  attention  has  been  given 
to  the  elementary  principles  of  projection,  in  order  to  lay  a 
very  thorough  foundation  for  later  w^ork. 

Exercises  intended  for  classroom  work  or  review  are  distributed 
through  the  text,  and  a  parallel  set,  intended  for  drafting-room 
w^ork,  is  given  in  the  appendix.  Each  set  of  exercises  covers  the 
same  ideas  and  in  the  same  order,  so  that  they  may  be  used 
interchangeably. 

In  preparing  the  course  on  which  this  text  is  based,  the 
writers  have  freely  used  the  standard  works,  particularly  those 
of  Pillet,  Warren,  AYillson,  Smith,  and  Anthony  and  Ashley. 

The  use  of  a  limited  number  of  models  has  been  found  advan- 
tageous. The  method  has  been  to  require  the  student  to  draw 
a  freehand  sketch,  in  plan  and  elevation,  from  the  model  and 
to  complete  his  working  drawing  from  the  sketch,  without  further 
reference  to  the  model. 


CONTENTS 

Chapter  Page 

I     Purpose  and  Scope .1 

II     Projection  of  Points .10 

III  Projection  of  Lines 20 

IV  Projection  of  Points  and  Lines  in  the  Second,  Third 

AND  Fourth  Quadrants  ......       42 

V  Plane  Figures  and  Solids    ......  53 

VI  Problems  Dealing  with  Points,  Lines,  and  Planes  66 

VII     Curved  Lines 119 

VIII     Curved  Surfaces .  153 

IX  Applications  for  Chapters  I  to  V      .         .         .         .  204 

X  Applications  for  Chapter  VI       ....         .  212 

XI  Applications  for  Chapters  VII  and  VIII  .         .         .  226 

XII  Shades  and  Shadows        .......  242 

XIII  Other  Methods  of  Projection      .....  262 

Appendix 275 


GENERAL  EXPLANATIONS  AND 
INSTRUCTIONS 

Notation.  The  system  of  indication  and  notation  shown  in 
the  accompanying  cuts  is  used  throughout  the  text.  Students 
are  expected  to  use  this  system  for  all  exercises.  It  should  be 
recognized,  however,  that  this  notation  is  merely  a  means  to  an 
end.  Some  system  is  needed  in  the  earlier  problems  in  order 
to  make  the  answers  definite  and  complete ;  in  the  later  prob- 
lems, a  good  system  simplifies  the  actual  processes  of  the 
solution. 

While  system  and  systems  are  valuable  aids,  they  should 
never  be  allowed  to  take  the  place  of  a  thorough  comprehension 
of  the  problem  and  a  visualization  of  the  facts.  The  system 
here  shown  is  that  followed  generally  throughout  the  text.  It 
has  been  changed  occasionally  to  suit  the  requirements  of  indi- 
vidual illustrations. 

Exercises.  The  problems  in  the  body  of  the  text  are  intended 
for  classroom  or  home  work.  In  general,  they  are  arranged  in 
groups,  each  group  treating  a  single  idea.  They  are  intended 
to  be  laid  out  on  8''xl0^''  paper,  the  standard  loose  leaf  note- 
book size,  with  margins,  etc.,  as  shown  on  page  xii. 

The  problems  given  in  the  appendix  cover  the  same  ground 
as  those  in  the  text,  but  are  intended  for  drafting-room  work. 
They  are  arranged  to  go  either  on  8"xl0^''  sheets,  or  on 
15''X22''  sheets  (standard  *' Imperial  "  size,  cut  in  two),  as 
shown  on  page  xii.  It  will  be  noticed  that  the  15''x22'' 
sheet  will  accommodate  four  of  the  smaller  drawings,  thus 
making  it  possible  to  use  the  larger  size  only,  if  desired. 

All  exercises  can  be  adapted  to  blackboard  work  by  changing 
the  unit  of  measure. 

ix 


X        GENERAL  EXPLANATIONS  AND  INSTRUCTIONS 


.Ij^, P   Isometric   of,  point,  line  and  plane. 
\      -  V 


Orthographic     projection    fn^ 
of  above   construction. 


Description  of  Quantities.  A  point  is  indicated  by  a  small 
(lower  case)  letter,  and  its  distances  from  the  tliree  planes  of 
projection  (profile,  horizontal,  and  vertical)  are  given   in  that 


GENERAL  EXPLANATIONS  AND  INSTRUCTIONS        xi 

NOTATION  AND  INDICATION 

THE  CO-ORDINATE   PLANES 

Horizontal    plane - H 

Vertical  ■•         V 

Profile  ••        P 

Auxiliary   horizontal  plane H' 

vertical         "        _    . V' 

profile  "        P' 

Origin    (Intersection  of  H.Vond  P) A 

Horizontal  axis -    _    .    . _    .  HA 

Vertical        ••         VA 

Profile  ••         - PA 

The  four  quadrants    . I.H.IU.IZ 

POINTS      (Smoll  letters) 

Point,  "q"  in  space --     a 

"    moved  to  new  psitions     .    -    _     .    .    _    .    .    a',  a" 
Projections  of  "a"  on  H.V  and  P        _______    a. c^a'* 

"  "    "n'    "   any  plane,    e.g.  RorV  _____    n.n^ 

"  "     moved  points    .    _    _    _    .    -    _    .    .    _    b',d' 

LINES 

y      Visible  Invisible  v 

Given  or  required     (projections)  _    _    _      a b 

Auxiliary   ___________      k 9 

Traces    ( piercing  points ) .    -    .     .    _    . «» «^^ 

Projecting    lines       ..-___--        ---  =  --- - 

Showing  movement  and  direction    .    -     d- -»- *d' 

PLANES     (Capitol  letters)  ^ 

.^  .  .     ,  H  ^^sM^^  A 

Iraces,  given   or  required  .   _   .   .    . 


H  — ■  ^^ A 

Auxiliary    plane  ^  ^^ 


xii       GENERAL  EXPLANATIONS  AND  INSTRL'CTIONS 


PLATE      LAYOUTS 


lot" 

-  lO* 


C^OO 


The  smaller  (8"Mok")  sheet 
i5  for  note -book  or  home 
work  .  The  larger  one(i5'V22') 
is  for  drafting-room  use  .  The 
subdivisions  of  the  latter  are 
the  same  size  as  the  small 
sheet,  making  the  problems 
adaptable   to  either. 


order  and  in  units  of  i".  Thus  a  (14,  6,  3)  indicates  that  the 
point  a  is  14  units  (1|")  from  P,  6  units  {f")  from  //,  and  3 
units  (f")  from  V.  When  the  point  described  is  in  a  quadrant 
other  than  the  first,  a  Roman  numeral  is  used  to  indicate  this 
fact ;  thus  b  (III,  17,  4,  9)  means  that  the  point  b  is  in  the  third 
quadrant,  17  units  from  P,  4  from  H,  and  9  from  I'. 


GENERAL  EXPLANATIONS  AND  INSTRUCTIONS     xiii 

A  line  is  described  by  any  two  of  its  points,  for  example  e 
(4,  7,  3)  /  (IV,  16,  3,  7) ;  or  by  one  point  and  its  direction.  A 
normal-case  plane  is  described  thus:  7^(17,  T''30°r,  T^'45° /), 
which  means  that  the  traces  of  the  plane  T  meet  on  HA,  17 
units  from  P,  the  horizontal  trace  making  an  angle  of  30° 
with  HAy  to  the  right,  while  the  vertical  trace  makes  an  angle 
of  45°  with  HA  to  the  left.     Special  cases  are  described  in  full. 

Presentation.  The  most  important  items  under  this  head 
are  accuracy  and  clearness.  It  is  recommended  that  the  solu- 
tions of  problems,  except  in  special  cases,  be  not  inked,  but  that 
clean  pencil  work  be  encouraged.  In  general,  indicate  all 
construction  lines  very  lightly,  and  given  or  required  lines  more 
strongly.  Distances,  angles,  etc.,  are  not  indicated  on  draw- 
ings unless  they  are  specified.  Be  careful  to  letter  each  point 
properly,  particularly  in  the  earlier  problems. 


:"/ 


DESCRIPTIVE  GEOMETRY 


CHAPTER  I 
PURPOSE   AND   SCOPE 

1.  Introduction.  When  several  persons  are  to  cooperate  in 
making  an  object  or  in  erecting  a  structure,  the  work  of  all 
must  be  controlled  by  a  preconceived  plan,  so  that  all  may 
work  inteUigently  toward  a  definite  end. 

For  very  simple  work,  the  plan  can  be  carried  in  the  mind 
and  the  workers  can  be  directed  by  word  of  mouth.  In  most 
cases,  however,  a  full  description  of  the  object  or  structure 
is  committed  to  paper.  In  this  way,  it  is  possible  to  foresee 
more  fully  all  complications  which  may  arise ;  a  greater  degree 
of  clearness  and  accuracy  is  possible  ;  and  a  record  may  be  kept 
for  reference  and  for  reproduction. 

Some  features  of  the  proposed  structure  can  be  described  most 
easily  in  a  written  specification.  Thus,  the  quality  of  materials 
and  the  details  of  workmanship  may  be  described  in  words. 
But  all  matters  pertaining  to  size,  shape,  disposition  of  parts, 
and  general  appearance  may  be  shown  much  more  readily  by 
means  of  drawings.  In  fact  many  parts  such  as  moldings, 
ornament,  etc.,  which  would  defy  a  description  in  words,  can  be 
explained  easily  by  a  small  and  simple  drawing. 

The  making  and  reading  of  these  descriptive  or  working 
drawings  is  based  on  a  line  language,  as  complete  in  its  way  as 
any  word  language.  Descriptive  Geometry  is  the  science  which 
underlies  this  line  language,  just  as  grammar  is  the  science  which 
underlies  word  language. 

1 


'2'.:  /'''A-;:.:  \'Pesg.riptive  geometry  [i,  §  i 

One  may  acquire  a  superficial  but  usable  knowledge  of  a 
foreign  language  without  understanding  its  grammar.  One  may 
also  be  able  to  make  useful  working  drawings  without  under- 
standing the  underlying  geometric  principles.  But  if  any  lan- 
guage, whether  of  w^ord  or  of  line,  is  to  be  used  with  confidence 
and  skill,  its  underlying  science  must  be  thoroughly  mastered. 

^Yorking  drawings  usually  represent  an  object  w^iich  is  to  be 
made  for  the  first  time.  Nothing  exactly  like  it  exists.  The 
person  making  the  drawing  must  therefore  depend  solely  on  his 
mental  image  of  the  object  that  is  to  be  made.  If  he  is  de- 
signing a  building,  he  must  be  able  to  see  his  phantom  structure 
from  many  points  of  view.  He  must  be  able  mentally  to  go 
through  and  around  it,  and  to  imagine  its  appearance  in  differ- 
ent lights  and  seasons. 

In  order  to  produce  anything  worth  w^hile  the  architect  or 
engineer  must  have  first  of  all  imagination  and  the  ability  to 
visualize.  But  he  must  also  have  some  means  of  recording  his 
ideas,  so  that  others  may  be  able  accurately  to  carry  them  out. 
The  study  of  descriptive  geometry  develops  the  power  of  visu- 
alization to  a  marked  degree.  Its  principles  furnish  the  basis  for 
recording  an  idea  in  the  form  of  accurate  and  complete  drawings. 
For  these  reasons  the  study  of  descriptive  geometry  is  taken 
up  early  in  schools  of  architecture  and  in  schools  of  engineering. 
When  any  study  has  two  aspects,  one  immediately  applicable 
to  real  problems,  the  other  more  or  less  abstract  (and  hence  in 
general  more  difficult),  there  is  a  strong  temptation  to  neglect 
the  abstract  in  favor  of  the  so-called  "  practical."  But  an  idea 
is  worth  more  than  its  bald  statement.  Learning  to  think  is 
more  important  than  learning  facts.  The  chief  value  of  de- 
scriptive geometry  lies  in  the  cultivation  of  the  imagination, 
and  not  in  the  acquisition  of  tricks  of  drawing,  or  of  methods 
for  the  solution  of  special  problems. 

Students  are  urged  to  keep  this  idea  in  mind  while  studying 
the  problems  which  follow.     No  problem  should  be  left  behind 


I,  §2] 


PURPOSE  AND  SCOPE 


until  there  is  formed  in  the  mind  a  clear  and  satisfactory  picture 
of  the  conditions  and  of  the  solution. 

2.  Kinds  of  Drawing.  The  fundamental  problem  of  all 
drawing  lies  in  representing  three-dimensional  objects  on  a  two- 
dimensional  sheet  of  paper.  Two  methods  are  commonly 
employed  in  making  such  representations.  One  method  is 
known  as  perspective  drawing,  the  other  as  projection  drawing. 


_-Oit 


Perspec+ive- 


FiG.  4 


In  perspective  drawing,  the  idea  is  to  produce  a  more  or 
less  exact  representation  of  the  object  as  it  appears  to  the  eye, 
including  the  well-known  foreshortening  effect  of  distance. 
This  foreshortening  (see  Fig.  4)  is  due  to  the  variation  between 
the  visual  angles  a  and  a'.  Its  effect  is  to  reduce  remote  parts 
and  to  crowd  the  details  there.  It  complicates  the  making  and 
reading  of  the  drawings.  Consequently,  perspective  drawings 
are  ordinarily  used  only  for  pictorial  purposes. 

The  other  method,  projection  drawing,  arbitrarily  eliminates 
the  natural  foreshortening,  so  that  all  parts  which  are  of  the 


DESCRIPTIVE  GEOMETRY 


[I,  §2 


Fig.  5 


Perspective 


same  size  are  so  represented  in  the  drawing.  This  destroys  the 
pictorial  effect,  but  it  makes  the  drawing  more  useful  for  struc- 
tural purposes.  Although  there  are  several  methods  of  pro- 
jection drawing,  only  the 
two  most  usual  ones  (ortho- 
graphic and  isometric),  will 
be  considered  here. 

Isometric  projection  (see 
Fig.  6)  shows  the  object  in 
all  three  dimensions  on  one 
drawing.  It  is  useful  to  a 
limited  extent  in  showing 
complicated  details  not 
otherwise  easily  explained.^ 
Orthographic  projection  (see  Fig.  7)  shows  but  two  di- 
mensions (one  face)  of  an  object  on  a  single  drawing.  At  least 
two,  usually  three,  and  often  more,  such  drawings  are  needed 
fully  to  describe  an 
object.  However, 
the  greater  clearness 
of  such  drawings 
and  the  rapidity 
and  ease  with  which 
they  can  be  made, 
render  them  far 
more  useful  than 
isometric  or  per- 
spective drawings. 
For  this  reason, 
orthographic  pro- 
jection is  used  for  the  most  part  in  this  book.  For  those  illus- 
trations in  which  a  pictorial  quality  is  desired,  isometric  draw- 

^  For  a  more  extended  treatment  of  isometric  projection  and  other  similar 
methods,  see  Chapter  XIII. 


Fig.  6 


I,  §3] 


PURPOSE  AND  SCOPE 


ings  have  been  used,  in  general,  though  CavaHer  projections 
or  perspective  drawings  have  been  substituted  in  a  few  cases. 

3.  The  Theory  of  Plan  Drawing  and  of  Building  from  Plans. ^ 
The  size  and  shape  of  any  three-dimensional  object  is  determined 
by  its  bounding  planes  or  surfaces,  and  these  in  turn  by  their 


Orthographic 
Fig.  7 


Projeciion 


bounding  lines.  Any  line  is  determined  by  any  two  of  its 
points.  Hence  the  size  and  shape  of  any  object  are  fixed  by  the 
interrelations  of  its  limiting  points.  For  example,  in  Fig.  7 
the  roof  surface  A  is  defined  b}^  the  lines  ab,  be,  cd,  and  da. 
These  lines  in  turn  are  determined  by  the  four  points  a,  b,  c,  and 
d.     These  four  points  then  limit  or  define  the  surface  A. 

In  starting  a  set  of  working  drawings,  the  draftsman  first 

^  ^  The  word  "plans"  is  sometimes  used  to  indicate  particular  projections 
of  an  object  (see  Fig.  7)  and  sometimes  to  indicate  the  whole  set  of  draw- 
ings.    Here  it  is  used  in  fhe  latter  sense. 


6  DESCRIPTIVE  GEOMETRY  [I,  §  3 

draws  some  lines  to  represent  certain  of  the  salient  lines  of  the 
proposed  structure  (such  as  the  axes  in  Fig.  7).  Then,  measuring 
from  these  lines  as  bases  of  reference,  he  locates  the  more 
important  points  and  lines  and  finally  the  details  of  the  pro- 
posed structure.  Proceeding  thus  from  point  to  point,  draw- 
ings are  finally  evolved  which  fully  describe  the  position  of  every 
point  and  line  with  reference  to  one  another  and  with  reference 
to  the  first  determined  axes.  AVith  these  drawings  as  a  guide,  the 
structure  they  represent  can  be  made  an\n^'here,  at  any  time. 

Since  the  representation  of  a  point  is  so  fundamental,  it  is 
worth  while  to  study  it  carefully  at  the  start.  The  purpose  of 
such  a  study  is  to  describe  a  method  by  which  the  location  of 
a  point,  and  from  this  the  location  of  other  points  and  lines, 
can  be  described  so  clearly  that  they  can  be  materialized  into 
a  perfectly  definite  structure. 

The  location  of  any  point  in  space  may  be  described  most 
readily  by  means  of  its  relations  to  three  fixed  planes  of  reference. 
For  example,  a  mountain  peak  may  be  located  by  its  latitude, 
its  longitude,  and  its  altitude.  A  building  may  be  located  by 
reference  to  two  of  the  lot  lines  and  to  the  curb  level.  The 
position  of  a  suspended  light  in  a  room  may  be  fixed  definitely 
by  stating  its  distance  from  the  end,  from  the  side  walls,  and 
from  the  floor.  In  the  latter  case,  if  the  distance  from  the  floor 
alone  is  given,  the  light  may  be  an^^'here  in  a  plane  parallel 
to  the  floor  and  at  the  given  height.  If  the  distances  from  the 
floor  and  from  one  wall  are  given,  the  light  may  be  anywhere  on  a 
line  parallel  to  the  floor  and  the  wall  But  when  its  distances 
from  all  three  planes  are  given,  its  location  is  definitely  fixed. 
The  walls  and  the  floor  form  three  fixed  planes  of  reference. 
Such  planes  are  called  the  datum  planes  for  locating  the  light. 

These  datum  planes,  or  planes  of  reference,  may  be  chosen 
at  will.  Ordinarily,  however,  each  plane  is  taken  perpendicular 
to  the  other  two  (as  in  the  cases  cited  above),  one  being  hori- 
zontal, and  the  other  two  vertical.     This  will  be  done  in  this 


I,  §3] 


PURPOSE  AND  SCOPE 


book  except  where  otherwise  stated.  The  object  to  be  drawn 
is  conceived  to  be  placed  within  the  solid  angle  formed  by 
these  planes.     Each  salient  point  of  the  object  is  then  projected 


H  =  the  horizontal 

plane  of  projection 

F=the  vertical  plane 
of  projection 

P  =  the  profile  plane 
of  projection 


''  =  horizontal  projection  of  a 

av  =  vertical  projection  of  a 
j^jQ_  g  aP  =  profile  projection  of  a  ^ 

aa'' =horizontal  projecting  line 
aa'^^ vertical  projecting  hne 
aaP= profile  projecting  line 


on  each  plane  bv  means  of  a  line  passing  through  the  point 
perpendicular  to^he  plane.  (See  Fig.  8.)  The  foot  of  such  a 
perpendicular  is  called  the  projection  of  the  point. 


8  DESCRIPTIVE  GEOMETRY  [I,  §  3 

When  a  sufficient  number  of  points  have  been  projected,  the 
projections  on  each  plane  are  connected  by  Hnes.  The  projec- 
tions of  the  object  thus  obtained  form  descriptive  drawings. 
Since  the  projecting  lines  to  each  one  of  the  planes  are  all 
parallel,  the  relative  positions  of  the  various  points  are  main- 
tained in  the  projections.  (Compare  and  contrast  Figs.  5  and 
7.)     Thus  foreshortening  and  distortion  are  avoided. 

It  follows  that  the  process  of  drawing  plans  is  essentially 
that  of  projecting  points  on  arbitrarily  selected  planes,  while 
the  process  of  building  consists  in  establishing  the  planes  of 
reference  as  given  by  the  plans  and  projecting  the  points,  in 
a  reverse  sense,  back  into  space. 

4.  Special  Applications.  All  of  the  various  methods  of 
drawing  in  a  more  or  less  pictorial  way,  such  as  perspective 
drawing,  isometric  drawing,  shades  and  shadows,  etc.,  are 
special  applications  of  descriptive  geometry. 

The  determination  of  the  actual  size  and  shape  of  the  individ- 
ual parts  of  a  complex  structure  (for  example,  the  cutting  of  the 
stones  for  a  vault) ,  so  that  each  may  fit  perfectly  into  its  proper 
place,  is  a  problem  that  becomes  increasingly  important  with  the 
present-day  tendency  toward  shop  production.  The  principles 
of  descriptive  geometry  are  a  sure  guide  for  all  such  work.  All 
the  rules,  formulas,  and  practice  of  pattern  making  depend 
ultimately  on  the  same  principles. 

In  the  following  chapters  the  more  important  theorems  of 
descriptive  geometry  will  be  presented  in  a  purely  abstract 
fashion,  and  then  some  of  them  will  be  applied  to  the  solution  of 
concrete  problems.  It  is  hoped  that  through  the  frequent 
cross  references,  the  student  will  come  to  realize  that  no  problem 
stands .  alone,  that  the  whole  subject  is  coherent,  and  that  it 
cannot  be  mastered  piecemeal. 

5.  Historical  Note.  As  in  the  case  of  most  sciences,  the 
main  ideas  of  descriptive  geometry  were  in  daily  use  long  be- 
fore  its  principles   were   formulated  into  a  separate   science. 


I,  §  5]  PURPOSE  AND  SCOPE  9 

The  early  architects,  engineers,  and  craftsmen  worked  from 
plans  that  were,  from  a  modern  point  of  view,  imperfect  and 
incomplete.  The  rules  and  methods  used  in  making  the  plans 
and  in  carrying  out  the  work  were  passed  on  from  generation  to 
generation  by  word  of  mouth,  chiefly  as  guild  secrets,  which  con- 
sisted largely  of  mere  rules  and  formulas. 

It  remained  for  the  celebrated  French  mathematician,  Gaspard 
Monge  (1746-1818),  to  formulate  into  a  distinct  science  the  prin- 
ciples already  in  use.  The  way  was  then  prepared  for  rapid, 
sure,  and  permanent  progress.  The  development  of  modern 
practice,  infinitely  more  efficient  and  more  flexible  than  the  old, 
was  thus  made  possible. 


CHAPTER  II 
PROJECTION   OF  POINTS 

6.  Introduction.  In  the  previous  chapter  it  was  shown,  by 
the  use  of  Fig.  8,  how  the  projections  of  a  given  body  may  be 
found.  Usually  it  is  easier  to  visualize  the  process  of  projec- 
tion when  it  is  applied  to  a  body  than  when  it  is  applied  to  a  single 
point.     This  chapter  is  devoted  to  a  study  of  point  projection. 


i7A=  horizontal  axis 
yA=  vertical  axis 
PA  =  profile  axis 


ew 


Fig.  9 


7.  Fundamentals  of  Point  Projection.  Figure  9  shows  the 
method  of  orthographic  projection  applied  to  a  single  point. 
It  should  be  carefully  studied  in  order  to  fix  thoroughly  in  the 
mind  a  few  fundamental  relations.     (See  footnote,  p.  7.) 

The  figure  shows  the  point  a  in  space,  projected  on  the  planes 
//,  V,  and  P.  The  points  a^,  a^,  and  a^  are,  respectively,  the 
horizontal,  vertical,  and  profile  projections  of  a.  If  now  the 
point  is  removed,  while  the  projections  are  retained,  the  point  a 

10 


II,  §  8]  PROJECTION  OF  POINTS  11 

itself  can  be  exactly  restored  at  any  time  to  its  original  position, 
since  that  position  is  recorded  by  the  projections. 

The  projections  of  a  point  are  thus  a  record  or  a  representation 
of  the  point  in  space.  Such  a  representation  of  a  point  is  quite 
as  distinct  from  the  point  itself  as  a  sign  board  is  distinct  from 
the  town  toward  which  it  points.  The  projections  of  a  point 
bear  exactly  the  same  relation  to  a  point  as  the  plans  of  a 
building  bear  to  the  building  itself.  Referring  again  to  Fig.  9, 
observe  that  the  point  a,  together  with  its  projections  a'",  a^,  and 
aP,  determine  a  right  parallelepiped.  It  is  obvious  that  the 
opposite  sides  are  equal.     Hence  the  distances  of 

a^  from  HA  and  PA  represent  the  distances  of  a  from  V  and  P, 
a-  from  HA  and  VA  represent  the  distances  of  a  from  H  and  P, 
aP  from  PA  and  VA  represent  the  distances  of  a  from  H  and  T^. 

In  other  words  the  location  of  a  point  with  respect  to  T"  and 
P  is  shown  on  H ;  its  location  with  respect  to  H  and  V,  on  P ; 
and  its  location  with  respect  to  H  and  P,  on  V.  Thus  no  single 
projection  can  locate  a  point  definitely  with  respect  to  all  three 
reference  planes.  (Compare  the  H  projection  of  a  and  a'  in 
Fig.  9.)  But  any  two  projections,  taken  together,  do  definitely 
locate  a  point  in  space,  since  the  two  taken  together  show  its 
location  with  respect  to  all  three  of  the  planes  of  reference. 

8.  Relation  to  Plan  and  Elevation.  ^Mien  a  solid  body  is  to 
be  drawn,  as  in  Fig.  8,  relations  similar  to  those  of  §  7  hold  true. 

The  V  projection  shows  relations  of  height  and  length. 
The  P  projection  shows  relations  of  height  and  width. 
The  //  projection  shows  relations  of  length  and  width. 

Since  any  two  of  the  projections  taken  together  show  points 
in  their  three-dimensional  relations,  it  is  quite  possible  to  show 
fully  the  size  and  shape  of  a  simple  body  by  projecting  it  upon 
only  two  planes.  "When  it  becomes  necessary  to  fix  the  position 
of  the  body  in  space,  however,  the  third  plane  of  reference  must 


12 


DESCRIPTIVE  GEOMETRY 


[11,  §  8 


be  used,  even  though  the  projection  on  this  third  plane  may  not 
actually  be  drawn.  Moreover,  when  a  body  becomes  more  com- 
plex and  many  points  have  coincident  projections  on  one  or 
another  plane,  as  in  Fig.  8,  it  becomes  necessary  to  draw  three  or 
more  projections  in  order  to  avoid  confusion.  Simplicity  and 
clearness  are  highly  desirable  in  this  work.     (See  Fig.  7.) 

The  horizontal  projection  of  a  body  is  commonly  called  the 
plan.  It  shows  the  flat  or  level  relations.  The  vertical  projec- 
tions are  called  elevations,  since  they  show  relations  of  height. 


Direc+ion    of    view 


Fig.  10 

9.  Representation  on  Paper.  To  outline  a  structure  of  any 
magnitude,  a  point  at  a  time,  on  three  mutually  perpendicular 
planes  of  projection  erected  in  space,  would  be  far  too  tedious 
and  quite  impracticable.  Methods  must  be  found  for  dealing 
with  points  in  wholesale  quantities,  a  line  at  a  time  or  a  plane 
at  a  time,  and  for  representing  them  on  flat  sheets  of  paper. 
We  shall  consider  first  the  problem  of  flattening  out  the  planes, 
so  as  to  show  the  projections  on  a  two-dimensional  surface. 

In  order  to  record  all  three  projections  on  a  single  flat  sheet, 
it  is  necessary  to  imagine  the  vertical  and  profile  planes  rotated 
about  their  axes,   until   they   lie   in  the  same  plane  with  the 


II,  §  10] 


PROJECTION   OF  POINTS 


13 


horizontal  plane  (//).  The  simplest  form  of  this  operation 
is  shown  in  Fig.  10,  which  shows  the  V  and  P  planes  separated 
along  VA  and  turned  away  from  each  other  into  the  plane  H. 

Let  us  now  see  what  happens  to  the  projections  of  a  point 
during  this  operation.  It  will  be  noted  that  as  V  revolves  on 
HA  as  an  axis,  the  T^  projection  of  a,  a'',  describes  an  arc  of  a 
circle  whose  plane  coincides  with  the  plane  of  a,  a^,  a^\  Hence 
the  line  a^'xa^  becomes  a  straight  line  perpendicular  to  HA. 
By  the  same  reasoning  the  line  a^za'' becomes  a  straight  line  per- 
pendicular to  A  P.  Figure  1).  shows  the  flat  drawing  which 
results  from  the  flattening  process  illustrated  in  Fig.  10. 

10.  Connections.  It  will  be  noted  that  the  erected  planes 
consist  of  three  flat  surfaces,  while  the  flat  drawing  is  divided 


V 

a". 

Ix 

H                           1                 A 
j'l 

z 

1                < 

-D 

Fig.  11 

into  four  parts  by  the  axis  lines.  No  projection  will  appear  on 
one  of  these  spaces  (VAV),  since  it  has  no  equivalent  in  the 
erected  planes.  Across  this  space  the  circular  arc  yy  is  drawn 
with  its  center  at  A.  The  significance  of  this  arc  is  frequently 
misunderstood.  It  signifies  that  before  the  planes  were  flattened, 
the  lines  a^'y  and  a^y  (Fig.  11)  touched  one  another  at  y,  i.e.  that 
the  two  points  y  represent  the  single  point  y  in  Fig.  10. 


14 


DESCRIPTIVE   GEOMETRY 


[II,  §  10 


A  line  drawn  at  an  angle  of  45°  to  VA  would  serve  the  same 
purpose  as  the  arc,  or  the  entire  line  a^^yya^  could  be  omitted. 
The  only  necessary  condition  is  that  a^z  =  a''x. 

11.  Alternative  Openings  for  the  Profile  Plane.  Let  the 
student  now  cut  out  a  card  or  sheet  of  stiff  paper  and  draw 


D     D 


H 


I 

I  ' 

1 ^ 


V 


A 


< 


Fig.   12 


on  it  the  projections  of  an  unsymmetric  object,  as  shown  in 
Fig.  12.  Now  let  the  card  be  placed  on  a  table,  face  up,  and 
let  the  parts  marked  V  and  P  be  bent  upward,  creasing  the 
card  along  the  lines  EA  and  AF.  There  will  now  be  three 
planes,  carrying  the  projections  of  the  given  object.  In  flat- 
tening out  these  planes,  V  is  simply  turned  backward  into 
U ;  while  P  may  be  turned  into  U  by  tipping  it  either  toward 
or  away  from  the  object  which  is  being  projected. 

Next  let  P  be  separated  from  E  by  cutting  along  AF,  let  it 
be  fastened  to  V  along  AV ,  and  let  the  solid  angle  be  erected 
as  before.  The  flattening  process  now  may  be  started  by  turn- 
ing F  OYi  AV  as  an  axis,  into  V .  Both  planes  are  then  folded 
backward  into  //.     In  this  operation,  the  P  plane  may  be  folded 


II,  §  12]  PROJECTION   OF  POINTS  15 

toward  or  away  from  the  object  to  be  drawn.  Thus  there  are 
four  ways  in  which  the  P  plane  may  be  opened. 

When  P  is  turned  into  V  and  then  both  planes  rotated  into 
H,  it  is  clear  that  the  two  elevations  are  seen  side  by  side. 
When  P  is  turned  at  once  into  H,  however,  the  plan  and  the  end 
elevation  are  shown  side  by  side,  while  the  front  and  the  end 
elevations  are  separated.  It  seems  more  logical,  and  is  actually 
easier  in  drawing,  to  have  the  two  elevations  closely  related. 

When  P  is  turned  towards  the  object,  the  end  elevations  will 
appear  as  if  viewed  through  P ;  whereas  when  P  is  turned  aioay 
from  the  object,  the  elevation  shows  the  object  reversed,  the  left- 
hand  side  appearing  at  the  right  of  the  drawing.  This  is  of  es- 
pecial importance  when  the  two  ends  are  not  alike,  so  that  both 
must  be  drawn. 

Thus,  of  the  four  possible  methods  of  rotating  the  P  plane, 
it  would  seem  best  to  use  the  one  giving  the  most  natural  result, 
viz. :  the  elevations  side  by  side  and  the  end  elevations  appearing 
as  if  viewed  through  the  plane.  In  other  words,  the  best 
method  is  first  to  fold  P  toward  the  object  into  V  and  then  to 
turn  V  (and  P)  backward  into  H. 

Although  the  preceding  method  will  be  used  most  frequently, 
it  is  often  desirable  to  use  some  one  of  the  others.  Consequently 
the  student  should  familiarize  himself  with  all  four  methods,  and 
with  placing  P  either  to  the  right  or  to  the  left  of  the  object.  The 
P  plane  may  be  used  on  either  one  or  on  both  sides  of  the  object, 
and  nearer  or  farther  from  it,  as  circumstances  may  require. 

12.  Normal  and  Special  Cases.  We  shall  classify  the  subject 
matter  so  that  the  majority  of  cases  fall  into  one  class,  while 
special  cases  that  occur  less  frequently  form  other  classes.  The 
solution  of  problems  is  more  rapid  and  more  accurate  with  such 
a  classification.  Thus,  in  trigonometry,  multiples  of  90°  are 
special  cases  of  angles. 

In  Fig.  9,  the  general  case  is  that  of  any  point  lying  within 
the  solid  angle  formed  by  H,  V,  and  P,  since  such  points  have 


16  DESCRIPTIVE   GEOMETRY  [II,  §  12 

the  same  general  properties  and  constitute  the  great  majority 
of  all  possible  points.^  We  shall  call  them  iionnal-case  points. 
But  if  a  point  lies  in  one  or  more  of  the  planes  of  projection, 
one  or  more  of  its  projections  will  have  special  characteristics. 
Such  a  point  will  be  called  a  special-case  point.^ 

The  ability  to  recognize  special  cases  and  to  make  use  of  their 
special  properties  is  often  the  key  to  the  best  solution  for  a 
problem.  The  auxiliary  points,  lines,  etc.,  that  are  used  in 
solving  problems  are  usually  special  cases,  and  their  special 
properties  make  them  particularly  useful. 

13.  Recapitulation  of  Chapter  II.  The  following  principles 
with  reference  to  the  projection  of  a  point  may  now  be  stated : 

1.  Three  planes  of  reference,  and  at  least  two  projections  are  neces- 
sary to  record  the  location  of  any  point. 

2.  To  show  the  size  and  shape  of  a  body,  hut  not  its  exact  location 
in  space,  one  of  the  planes  of  projection  may  he  omitted. 

3.  The  projections  of  a  point  on  any  two  planes  having  a  common 
axis  lie  in  the  same  straight  line  perpendicular  to  that  axis. 

4.  The  vertical  projection  of  a  point  is  as  far  from  HA  as  the 
actual  point  in  space  is  from  H. 

5.  The  horizontal  projection  of  a  point  is  as  far  from  HA  as 
the  actual  point  in  space  is  from  V. 

6.  The  vertical  and  horizontal  projections  of  a  point  are  as  far 
from  AV  and  AP  as  the  actual  point  in  space  is  from  P. 

7.  The  profile  projection  is  as  far  from  AP  as  the  point  is  from 
H,  and  as  far  from  AV  as  the  point  is  from  V. 

1  It  may  occur  to  the  student  that  many  points  in  space  may  lie  wholly 
outside  the  angle  between  the  planes  of  projection.  Such  cases  are  dis- 
cussed in  Chapter  IV. 

2  Undoubtedly  some  difficulty  will  be  experienced  in  the  use  of  these 
terms  when  applied  to  lines  and  planes.  No  two  persons  will  classify  them 
in  quite  the  same  manner.  It  is  felt,  however,  that  the  use  of  these  terms 
permits  a  most  desirable,  though  somewhat  vague,  grouping  of  the  subject 
matter,  which  will  be  found  to  have  justified  itself  in  the  long  run. 


II,  §  13] 


PROJECTION  OF  POINTS 


17 


8,  //  any  two  projections  of  a  jjoint  are  known,  the  third  can 
be  established.     (Result  of  4,  5,  6,  and  7,  above.) 

9.  Two  projections  of  a  point  may  be  assumed  at  random, 
within  the  restrictions  laid  down  in  3  above.  For  there  is  some 
point  in  space  corresponding  to  any  two  projections  that  may 
be  selected. 

PRELIMINARY  WRITTEN  EXERCISES 

1.  Prove  that  two  projections  of  a  point  lie  on  the  same  straight 
line  perpendicular  to  the  axis, 

2.  Prove  that  the  projections  of  a  point  are  at  the  same  distances 
from  the  axis  as  the  actual  point  in  space  is  from  the  planes  of  reference. 

3.  Describe  the  special  cases  of  points  that  may  occur,  and  tell  the 
distinguishing  characteristics  of  their  various  projections. 

4.  Describe  in  writing  the  following  points.  (See  the  description 
of  quantities,  p.  x.) 


-ic: 


d- 


d^ 


b^ 


Fig.  13 


V 

''h^   A 


5.    Describe  in  writing  the  following  points. 


V 


■,d^ 


U»' 


Id'' 


:<-h 


xqh 


FiQ.  14 


18 


DESCRIPTIVE   GEOMETRY 


[II,  §  13 


EXERCISE  SHEET  I 

[Note,  For  the  typical  layout  for  exercise  sheets,  applying  to  all 
exercises  in  the  text,  except  a  few  which  are  separately  described,  see 
p.  xii  of  the  general  explanations.] 

Take  P  at  the  right,  20  from  the  right  border  line,  turned  away 
from  the  given  points,  into  F.  Draw  all  three  projections  of  the 
following  points. 


Point 

Distance  from 

Point 

Distance  from 

P 

H 

V 

P 

H 

V 

a 

55 

15 

19 

/ 

27 

6 

8 

b 

49 

6 

0 

g 

20 

0 

0 

c 

43 

9 

2 

h 

15 

3 

8 

d 

36 

0 

14 

i 

9 

9 

7 

e 

32 

12 

12 

3 

3 

12 

12 

EXERCISE    SHEET   II 

Take  P  at  the  left,  4  from  the  border  line,  turned  toward  the  given 
points,  into  H.     Draw  all  three  projections  of  the  following  points. 


Point 

Distance  from 

Point 

Distance  from 

P 

H 

V 

P 

H 

V 

a 
b 
c 
d 
e 

16 
23 
30 
36 
42 

11 
8 

10 
0 
4 

5 
15 
10 
20 
10 

/ 

g 
h 

i 
J 

48 
53 
60 
65 

72 

0 

11 

2 
11 
10 

0 
0 
2 

20 
10 

II,  §  13]  PROJECTION  OF  POINTS  19 


EXERCISE   SHEET  III 

Take  P  at  the  left,  4  from  the  border  line,  turned  toward  the  given 
points,  into  V.     Draw  all  three  projections  of  the  following  points. 

a  (12,  8,  10)  d  (38,  13,  12)  g  (34,  0,  0)  j  (72,  13,  12) 
h  (19,23,  6)  e  (46,  8,  0)  h  (60,11,  8)  k  (50,  0,15) 
c     (28,  17,  19)       /     (53,  19,    4)        i     (66,    5,    8)        I     {  0,  10, 18) 


EXERCISE   SHEET  IV 

Take  P  at  the  right,  20  from  the  border  line,  turned  away  from  the 
given  points,  into  H.  Draw  all  three  projections  of  the  following 
points. 

1.  a  (55,  17,  9).     Move  a  9  to  the  right.     Call  the  new  position  a'. 

2.  6  (41,  6,  14).  Move  h  upward  until  it  Ues  in  the  same  horizontal 
plane  as  a. 

3.  c  is  34  from  P,  10  from  H,  and  lies  on  a  plane  passing  through 
HA  and  bisecting  the  dihedral  angle  between  H  and  V. 

4.  d  lies  in  the  same  horizontal  plane  as  c,  but  is  9  nearer  P  and  3 
farther  from  V.  Keeping  d  at  the  same  distance  from  H,  swing  it 
about  c  as  a  center  until  it  lies  in  the  same  line  perpendicular  to  V  as 
does  c. 

5.  e  is  in  a  straight  line  parallel  to  HA  with  d  but  at  19  from  P. 
Rotate  it  about  its  o\vn  H  projection  till  it  lies  in  H,  keeping  it  in  a 
plane  parallel  to  V. 

6.  /  lies  on  HA,  3  from  P.  Move  /  upward  14.  Call  this  position 
/'.  Move  the  point  forward  {i.e.  toward  the  observer)  until  it  is  5  in 
front  of  V.  Call  this  position  /".  Now  move  it  again,  this  time  5 
to  the  left,  and  mark  this  position  /'". 


CHAPTER   III 


PROJECTION   OF  LINES 

14.    Generation   of   a   Straight   Line.     If   any   point,   as   a, 
Fig.  15,  is  moved  to  a  new  position,  a' ,  the  projections  also  move 

to  new  positions,  as  shown. 
During  its  motion,  the  point  a 
has  generated  the  Une  aa' .  Simi- 
larly the  motions  of  the  projec- 
tions have  generated  the  lines 
a'a''"  and  a^a'^.  Also,  the  pro- 
jecting lines  {aa^  and  aoF)  of  the 
point  a  have  generated  the  pro- 
jecting planes  {aa^a'^  a'  and 
(wya'"  a')  of  the  line  aa' . 

15.  Straight  Line  Determined 
by  Two  Points.  One  and  only 
one  straight  Hne  can  be  drawn  through  two  points,  i.e.  any  two 
points  determine  a  straight  line.  Hence  the  projections  of  any 
two  points  determine  the 
projections  of  a  line. 

The  two  points  that  de- 
termine a  line  do  not  neces- 
sarily limit  the  line.  They 
merely  fix  its  position  in 
space.     The  line  is  thought  ^\^ 

of  as  extending  indefinitely 
in  both  directions. 

16.    Assuming  a  Line.     Any  two  lines  drawn  on  a  sheet  of 
paper,  one  above  and  the  other  below   HA  will  represent  the 

20 


Fig.  15 


H- 


A 


Fig.   16 


Ill,  §  17] 


PROJECTION   OF  LINES 


21 


projections  of  some  line  in  space.  This  fact  is  readily  seen  by 
reference  to  Fig.  16.  Let  d'h^  and  ay¥  be  any  two  lines  drawn 
as  above.  Consider  now 
Fig.  17,  where  these  pro- 
jections are  shown  in  an 
isometric  drawing. 
Through  a^h"^  pass  a  plane 
perpendicular  to  V,  and 
through  a^h^  a  plane  per- 
pendicular to  H.  These 
planes  will  intersect.  The 
line  of  intersection,  ah,  is 
the  line  whose  projections 
are  a^h"^  and  aJ'h^. 

17.    Length  and  Slope. 
It  is  evident  from  Fig.  15 

that  when  a  line  is  parallel  to  H  and  V,  each  of  its  projections 
is  equal  in  length  to  the  line,  and  is  parallel  to  HA. 

Figure  18  shows  a  line  ah  which 
is  parallel  to  V  but  inclined  to  H. 
It  is  evident  that  the  V  projection 
is  equal  in  length  to  the  given  line 
ah,  and  that  the  length  of  the  H 
projection  is  less  than  the  length 
of  the  line.  Moreover,  the  angle  6 
is  equal  to  the  angle  0.  Now 
imagine  the  line  ah  to  be  rotated  in 
the  plane  aa^h^h,  using  a  as  a  center. 
During  the  rotation,  the  length  of 
the  V  projection,  a^h^,  will  remain 
constant  and  equal  to  ah.  The  length  of  the  H  projection 
will  vary  with  the  cosine  of  the  varying  angle.  The  limits 
of  this  variation  are,  when  0=0°,  a^h^—ahj  and  when  0  =  90°, 
a^h^=0. 


22 


DESCRIPTIVE  GEOAIETRY 


[III,  §  17 


Figure  19  shows  a  line  which  is  inclined  to  both  planes  of 
projection.  It  will  be  evident  immediately  that  the  projections 
are  each  shorter  than  the  line  itself,  and  that  the  angles  a''  and 
0!"  between  the  projections  and  HA  are  greater  than  the  angles 


o6=a  line  in  space 
a^6^= horizontal  projection 
a''6''  =  vertical  projection 
aP6P  =  profile  projection 


Fig.   19 


abb^a^,  ahh^'a'',  and  abbPaP 
are  the  horizontal,  verti- 
cal, and  profile,  projecting 
planes 


a  and  /3,  which  the  line  makes  with  the  planes  of  projection. 
18.  Normal  and  Special  Cases.  The  previous  article  will 
suggest  the  natural  separation  between  the  normal  and  the 
special  cases  of  lines.  All  lines  not  parallel  to  one  or  more  of 
the  planes  of  projection  can  be  considered  as  normal  cases. 
Among  the  lines  parallel  to  one  or  more  of  the  planes  of  pro- 
jection a  number  of  special  cases  can  be  distinguished. 


Ill,  §  20]  PROJECTION  OF  LINES  23 

19.  Method  of  Study.  An  opened  book  cover  may  be  used 
to  represent  the  planes  of  projection.  Place  the  book  on  the 
table  with  its  binding  turned  away  from  the  observer.  Open 
the  top  cover  to  a  vertical  position.  Use  a  pencil  to  represent 
a  line,  and  use  the  book  and  cover  to  represent  H  and  V. 

20.  General  Theorems.  The  following  simple  theorems  are 
merely  stated,  the  proofs  being  left  as  exercises  for  the  student. 

1.  One  projection  does  not  determine  a  line. 

2.  Two  projections  definitely  determine  a  line. 

3.  If  two  projections  of  a  line  are  known,  the  third  can  he  found. 

4.  Any  two  lines  draum  at  random,  on  different  planes  of 
projection,  represent  the  projections  of  some  line  in  space. 

Compare  the  above  theorems  with  numbers  1,  8,  and  9  of  §  13. 

5.  When  a  line  is  parallel  to  V,  (a)  its  V  projection  is  equal 
in  length  to  the  line  itself,  (b)  the  angle  between  the  V  projection 
and  HA  is  equal  to  the  angle  between  the  line  and  H. 

6.  When  a  line  is  perpendicidar  to  V,  (a)  its  V  projection  is  a 
point,  (b)  the  projection  on  H  is  the  same  length  as  the  lijie  and 
perpendicular  to  HA. 

State  theorems  similar  to  5  and  6  for  the  planes  H  and  P. 

[Note.  A  line  which  is  perpendicular  to  any  plane  of  projection 
is  necessarily  parallel  to  the  other  two  planes,  if,  as  is  usual,  the  planes 
of  projection  are  perpendicular  to  each  other.] 

7.  When  a  line  is  not  parallel  to  a  plane  of  projection  its  true 
length  and  slope  are  not  shown  by  its  projections. 

8.  The  projection  of  a  line  is  never  longer  than  the  line  itself. 

9.  The  angle  between  the  V  projection  of  a  line  and  HA  is 
never  less  than  the  angle  between  the  line  and  H. 

EXERCISE   SHEET   V 

Take  P  at  the  right,  20  from  the  border,  and  opened  away  from  the 
given  lines,  into  V.     Draw  the  projections  of  the  following  lines. 

1.  Line  ah :  a  (54,  5,  9) ;   h  (44,  8,  2). 

2.  Line  cd :  c  (40,  12,  7) ;  d  is  nearer  to  V  and  H  than  is  c,  and 
31  from  P. 


24 


DESCRIPTIVE   GEOMETRY 


[III,  §  20 


3.  Line  cj :  e,  on  V,  28  from  P ;  /,  on  H,  20  from  P. 

4.  Line  ^/;  :   perpendicular  to,  but  not  touching,  V ;    19  above  H ', 
and  17  from  P. 

5.  Line  i; :  lies  on  //  and  runs  from  i  (14,  0,  10)  backward  and  to 
the  right,  to  j,  which  is  8  from  P . 

6.  Line  kl :   parallel  to,  and  4  from,  P  ]    k  m  nearer  V  and  farther 
from  H  than  is  I. 

21.    Visualizing  a  Line.     The  student  now  should  have  gained 
some  facility  in  visualizing  a  line.     Let  him  then  try  to  describe 

^  a  line  whose  projections  are 

given. 

Figure  20  shows  the  pro- 
jections of  a  line  ah  which 

H 1 ■ !  A     rises  and   recedes   from   V, 

as  it  passes  from  left  to 
right.  If  the  position  of 
a  line  is  not  readily  grasped 
from  its  projections,  each 
projection  should  be  considered  separately.  By  taking  succes- 
sive points  along  the  H  projection,  and  comparing  their  relative 
distances  from  HA,  the  slope  of  the  actual  line  with  regard  to 
V  can  be  determined.  The  same  process  with  the  F  projection 
gives  the  slope  of  the  actual  line  with  regard  to  H.  After  a 
little  practice  the  projections  of  a  line  will  suggest  its  position 
and  its  slope  with  regard  to  the  planes  of  projection. 

EXERCISE   VI 
Describe  in  writing  the  following  Hnes. 


Fig.  20 


Ill,  §  23]  PROJECTIOX  OF  LINES  25 

22.  Limiting  Projections  of  a  Plane  Figure.  Another  exer- 
cise of  value  consists  in  studying  the  possible  variations  in  the 
projections  of  plane  figures.  Try  a  square.  Place  it  in  various 
positions  with  respect  to  H,  V,  and  P,  and  note  the  projections 
in  the  limiting  cases  and  the  law  of  variation  between  the  limits. 

EXERCISE    SHEET   VII 

Take  P  at  the  right,  18  from  the  border  line  and  turned  away  from 
the  objects,  into  V.  Profile  projections  are  required  for  #3  only.  Use 
notation  carefully. 

(1)  Draw  the  H  and  T'  projections  of  a  square,  abed,  10  on  a  side, 
lying  in  a  plane  13  above  H,  and  with  the  back  left-hand  corner  at  a 
(55,  13,  4).  The  side  ah  is  parallel  to  V.  Rotate  the  figure  toward 
H  on  ad  as  an  axis,  until  it  lies  in  a  vertical  plane.  Draw  its  projec- 
tions in  this  position.  Now  with  ah'  as  an  axis,  swing  the  square 
through  45°  to  the  right  and  toward  V,  and  draw  its  projections  in 
this  position. 

(2)  The  center  of  a  circle,  12  in  diameter,  is  at  o  (30,  10,  8).  The 
plane  of  the  circle  is  parallel  to  V.  The  horizontal  diameter  is  eg,  and 
the  vertical  diameter  is  fh.  Rotate  the  figure  on  eg,  the  top  traveling 
away  from  T',  until  it  hes  in  a  horizontal  plane.  Draw  its  projections 
in  this  position. 

(3)  (Use  the  profile  plane.)  With  the  line  j  (15,  4,  5),  /:  (4,  4,  5) 
as  a  base,  construct  an  equilateral  triangle  whose  plane  is  perpendicular 
to  both  V  and  P.  On  jk  as  an  axis,  swing  the  triangle  upward  until 
it  is  parallel  to  T'.  Draw  the  projections  in  this  position  and  in  two 
intermediate  positions.     Use  a  different  kind  of  lino  for  each  position. 

23.  True  Length  of  a  Line.  In  §  17  it  was  pointed  out  that 
the  projections  of  a  normal-case  line  do  not  show  either  the  true 
length  of  the  line  in  space  or  its  true  inclination  with  respect  to 
the  planes  of  projection.  Ordinary  plans  and  elevations  make 
frequent  use  of  just  such  lines  to  indicate  structural  parts.  Thus, 
in  Fig.  199  a,  p.  219,  the  line  ac  indicates  the  hip  line  of  a  four- 
pitched  roof.  It  is  only  after  the  true  length  and  the  slope  of 
such  members  are  known  that  the  piece  can  be  made.  Hence 
special  methods  are  devised  for  determining  these  quantities. 

Two  general  methods  are  used  for  determining  the  true  lengths 


26 


DESCRIPTIVE   GEOMETRY 


[III,  §  23 


Ill,  §  24]  PROJECTION   OF  LINES  27 

and  the  slopes  of  normal-case  lines.  In  each  method  the 
fundamental  idea  is  to  arbitrarily  reduce  the  normal-case 
line  to  a  special  case  (parallel  to  a  plane  of  projection),  and  to 
find  its  projections  under  the  altered  conditions. 

24.  First  Method  for  Finding  True  Length.  The  first  method 
consists  in  swinging  the  line  about  until  it  is  parallel  to  one  of 
the  planes  of  projection.  But  since  on  paper  we  deal  w^ith  the 
projections  of  the  line  and  not  with  the  line  itself,  it  is  necessary 
to  have  a  definite  method  of  procedure  for  this  operation. 

The  general  method  can  be  explained  best  by  reference  to  a 
definite  case.  In  Fig.  22  a,  the  normal-case  line  ab  is  to  be  swung 
around  into  a  plane  parallel  to  T  ,  let  us  say.  In  order  to  define 
the  motion  of  the  line  and  to  follow  it  in  projection,  let  the  lower 
end  a  be  kept  in  its  original  position  throughout  the  operation, 
and  let  the  angle  a,  which  the  line  makes  with  H,  be  maintained. 

If  the  line  is  revolved  in  this  manner  through  360°,  each  of  its 
various  positions  will  correspond  to  an  element  of  a  cone  whose 
apex  is  at  a  and  whose  base  is  parallel  to  and  at  the  distance 
bb^  from  H.  AYhile  the  line  ab  is  thus  generating  a  cone  in 
space,  the  horizontal  projection  of  b  is  generating  on  H  sl  circle 
whose  center  is  at  a^  and  whose  radius  is  equal  to  a^b^,  the 
horizontal  projection  of  ab.  Indeed,  this  circle  is  the  horizontal 
projection  of  the  base  of  the  cone.  At  the  same  time,  the  vertical 
projection  of  b  is  moving  in  a  line  parallel  to  HA,  since  b  is 
always  at  the  same  distance  from  H. 

It  will  be  seen  that  for  every  position  of  ab  the  length  of  the 
horizontal  projection  is  the  same,  while  that  of  the  vertical  pro- 
jection varies  as  the  angle  between  ab  and  V  varies.  Now  a^b^ 
never  shows  the  true  length  of  ab  ;  but  there  are  two  positions  of 
a^'b""  which  do  show  it,  viz.,  those  which  correspond  to  ab  at  the 
particular  instant  when  it  is  parallel  to  V.  These  positions  cor- 
respond to  the  elements  of  sight  of  the  generated  cone.  In  Fig. 
22  b,  one  of  the  two  possible  special-case  positions  of  ab  is  marked 
ab'.     The  V  projection  of  ab'  {a^'b'^')  is  the  true  length  of  ab. 


28 


DESCRIPTIVE   GEOMETRY 


[HI,  §  24 


Fig.  23 


Turning  now  to  Fig.  23,  the  operation  is  repeated  in  projec- 
tion. The  //  and  V  projections  of  ab  are  given  and  it  is  re- 
quired to  find  the  true  length  of  ab  and  the  true  angle  which  it 

makes  with  H.  A  short  arc  is 
swung  from  a^  as  a  center,  using  a 
radius  equal  to  a'^b^.  This  arc 
represents  a  portion  of  the  cir- 
cumference of  the  base  of  the  cone 
in  horizontal  projection.  Next  a 
line  {a^b'^)  is  drawn  through  a^ 
parallel  to  HA.  This  line  repre- 
sents the  horizontal  projection  of 
ab  when  it  is  parallel  to  V.  It  is 
necessary  now  to  pass  to  the  V 
projection.  A  line  (b^x)  is  drawn 
through  the  V  projection  of  b,  parallel  to  HA.  This  line  repre- 
sents the  V  projection  of  the  base  of  the  cone.  The  F  projection 
of  the  point  b^  will  be  found  somewhere  on  this  line.  x\lso  it  will 
be  directly  above  the  H  projection  (6'^)  already  found  (3,  §  13). 
Hence,  if  a  line  is  drawn  through  b'^.  perpendicular  to  HA,  its 
intersection  with  b^x,  (6''),  will  determine 
the  V  projection  of  the  point  b  when  the 
line  ab  is  parallel  to  V.  Connecting  this 
point  with  a''  we  have  the  line  which  is 
the  V  projection  of  ab  w^hen  it  is  parallel 
to  V,  and  which  shows  the  true  length  of 
ab.  The  angle  a  is,  of  course,  the  real 
angle  })etween  ab  and  //. 

25.  The  True  Angle  with  V.  The  true 
angle  that  a  line  makes  with  V  may  be 
found,  if  required,  by  swinging   the   line 

into  a  plane  parallel  to    //,  instead  of  parallel  to   F,  as  was 
done  in'  §  24.     Such  a  case  is  shown  in  Fig.  24. 

Here  the  point  b  is  kept  stationary  while  a  is  rotated  in  a 


Fig.  24 


Ill,  §  26] 


PROJECTIOX   OF   LINES 


29 


plane  parallel  to  V.  The  generated  cone  in  this  case  has  its  base 
parallel  and  close  to  V,  while  the  apex  (b)  is  farther  from  V. 
The  line  b^a'^  shows  the  true  length 
and  the  angle  </>  is  the  true  angle  with  T^. 
26.  Variations  of  Preceding  Methods. 
In  either  of  the  above  cases,  either  end 
of  the  line  might  have  been  chosen  as 
the  stationary  end.  If  the  true  length 
only  is  wanted,  the  angles  not  being  re- 
quired, either  method  may  be  used. 

Figure  25  shows  the  above  variations 
on  the  general  method  applied  to  a 
single  case.  The  first  operation  con- 
sists in  swinging  the  line  parallel  to  V, 
keeping  the  end  a  stationary.  This  is  shown  in  Fig.  25  in  solid 
lines.     The  second  operation  consists  in  swinging  ab  parallel  to 

H,  using  b  as  the  still 
point.  The  true  lengths 
found  by  any  two  meth- 
ods should  agree.  The 
sum  of  the  angles  a  and 
<f>  never  exceeds  90°.^ 

1  The  proof  is  as  follows. 
(See  Fig.  26.)  Let  a'-'l^  be 
any  line.  The  angles  1  and 
2  are  the  angles  which  the 
given  line  makes  with  H  and 
T^  respectively.  To  prove  that 
1  +  2  cannot  exceed  90°.  In 
the  right  triangle  a'^a'^b",  the 
side  a^'a''  <  a'-'h".  The  right 
triangles  a^'a^b^  and  a'^b^b^ 
have  a  common  hypotenuse  ; 
hence  the  one  having  the 
greater  base  will  contain  the 
smaller  angle  adjacent  to  the 
base.  Thus  2  <  3.  But.  1  +  3  =  90°,  therefore  1  +  2  <  90°.  When  ab  is 
parallel  to  P,  a^fe"  coincides  with  a'"a^.  In  that  case  the  angles  2  and  3  are 
identical  and  1  +  2  =  90°.     But  1  +  2  cannot  exceed  90°. 


Fig.  26 


30 


DESCRIPTIVE   GEOMETRY 


[III,  §  27 


27.  Second  Method  for  Finding  True  Length.  The  second 
method,  as  well  as  the  first,  depends  on  having  the  line  parallel 
to  a  plane  of  projection.  It  differs  from  the  first  method,  however, 
in  that  the  plane  is  brought  to  the  line,  rather  than  the  line  to 
the  plane. 

The  student  is  already  familiar  with  the  idea  of  projecting  a 
point  on  three  planes.     (See  Chapter  II,  especially  3,  8,  §  13.) 


When     the     projection     of    a 
point    15    found    on    a     new 
plane  .  the    nofahon    corres- 
ponds    to    that    of    the 
plane 


y/hen    using    V 


When    using   V. 


Fig.  27 


The  only  extension  of  that  idea  which  is  required  here  is  that 
the  third  projection  may  be  made  on  a  plane  perpendicular  to 
//,  but  not  necessarily  perpendicular  to  V.  In  fact,  the  new 
plane  may  be  chosen  parallel  to  any  given  line,  thus  making  that 
line  a  special-case  line  with  reference  to  the  new  plane. 

The  operation  described  above  is  illustrated  in  Fig.  27.  The 
line  ab  and  its  projections,  a^b^  and  a^'fe^,  are  given.  It  is  required 
to  find  the  true  length  and  the  angle  of  inclination  to  //.     A  new 


Ill,  §  27] 


PROJECTION  OF  LINES 


31 


plane  (F')  is  chosen  perpendicular  to  H,  and  parallel  to  ah. 
The  line  is  projected  on  this  plane  in  the  usual  manner,  and  the 
plane  folded  into  H.  The  projections  thus  found  will  show  the 
required  facts. 

Figure  28  shows  the  same  operation  in  projection.  Starting 
with  the  given  H  and  T^  projections  {a!^h^  and  a'-'h^'),  the  auxiliary 
axis  H' A'  is  drawn  parallel  to  a^h^  to  represent  the  axis  line  be- 
tween the  new  plane  and  H.  The  projections  of  a  and  h  on  this 
new  plane  will  lie  on  lines  passing  tlirough  a^  and  h^  and  per- 
pendicular to  H'A'  (3,  §  13). 
Let  two  such  lines  {a^r  and 
6^5)  be  drawn,  and  let  them 
be  of  indefinite  length.  The 
height  of  the  points  a  and  h 
(in  space)  from  H  will  govern 
the  distances  of  the  new  {V) 
projections  from  H'A'.  (Com- 
pare, in  Fig.  27,  bb^,  b'y,  and 
b^'y').  If  then  we  measure 
off  x'a^'  and  y'b^'  equal  to  xa" 
and  yb^  respectively,  we  will 
have  determined  the  V  pro- 
jections of  the  points  a  and  6, 
and  hence  of  the  line  ab.  This  projection  shows  the  true  length 
of  the  line  (a'"'6'0  and  the  true  angle  of  inclination  to  H,  0,  as 
required. 

Since  three  planes  of  projection  are  used  in  the  operation, 
three  projections  result.  Two  are  elevations  and  one  is  a  plan. 
The  plan  is  common  to  the  two  elevations.  The  process  of  mak- 
ing the  new  projection  is  sometimes  more  easily  visualized  if 
the  student  turns  the  drawing  about  until  H'A'  is  horizontal, 
thus  placing  the  new  plane  in  the  position  that  V  occupied  for- 
merly.    Note  the  arrows  in  Fig.  27. 

Figure  29  shows  the  same  method  applied  to  finding  the  true 


32 


DESCRIPTIVE  GEOMETRY 


[III,  §  27 


length  and  the  angle  of  inclination  to  V.     In  this  case  the  new 
plane  must  be  taken  perpendicular  to  V  and  parallel  to  the  given 

line,  so  as  to  show  the  true  angle 
with  V.  Of  course  the  true  lengths 
as  determined  in  Figs.  28  and  29 
should  agree. 

28.  Comparison  of  the  First  and 
Second  Methods  for  Finding  True 
Lengths.  When  the  methods  of 
§§24  and  27  are  carefully  compared, 
it  will  be  noticed  that  the  second 
method  is  simpler,  both  in  concep- 
tion and  in  execution.  However, 
both  of  them  depend  on  the  same 
idea,  i.e.  that  of  arbitrarily  altering 
the  given  conditions  by  moving 
either  the  line  or  a  plane  of  projec- 
tion so  that  the  conditions  become 
those  of  a  special  case.  Sometimes 
one  method  is  more  desirable,  and  sometimes  the  other.  The 
student  should  be  able  to  use  either  method  with  facility. 

The  principle  of  the  second  method,  viz.,  that  it  is  always  possi- 
ble, given  a  plan  and  elevation,  to  construct  a  new  elevation 
projected  from  a  new  point  of  view,  is  used  in  many  ways  in 
descriptive  geometry  and  in  its  application  to  working  drawings. 


Fig.  29 


EXERCISE   SHEET  VIII 

In  Exs.  1-5^  take  P  at  the  right,  16  from  the  border  line  and  turned 
away  from  the  objects,  into  V.  Draw  the  profile  projections  in  prob- 
lems 4  and  5  only.     In  Exs.  1-4,  use  the  first  method. 

1.  Given  a  (60,  6,  7)  and  b  (51,  13,  11),  find  the  true  length  of  ab 
and  the  angle  it  makes  with  H. 

2.  Given  c  (44,  7,  2)  and  d  (39,  11,  6),  find  the  true  length  of  cd  and 
the  true  angle  it  makes  with  V. 


Ill,  §  29]  PROJECTION  OF  LINES  33 

3.  Given  e  (33,  10,  14)  and/  (25,  4,  6),  find  the  true  length  of  ej  and 
the  true  angle  it  makes  with  V,  by  revolving  it  about  its  center  point  g. 

4.  Given  h  (22,  12,  4)  and  i  (14,  22,  11),  find  the  true  length  of  hi 
and  the  angles  it  makes  with  H  and  V,  by  swinging  the  line  about  i  as 
a  center  until  it  is  parallel  to  P. 

5.  Given  y  (11,  10,  6),  k  (11,  2,  6),  I  (3,  3,  12),  m  (5,  8,  10^),  draw 
all  three  projections  of  the  quadrilateral  jkhn  and  find  its  true  shape. 

EXERCISE  SHEET  IX 

Take  P  at  the  right,  16  from  the  border  line  and  turned  away  from 
the  objects,  into  H.     Use  P  in  problem  5  only.     Use  the  second  method. 

1.  Given  a  (60,  5,  7)  and  h  (52,  11,  5),  find  the  true  length  of 
ab  and  the  angle  it  makes  with  H. 

2.  Given  c  (46,  7,  6)  and  d  (38,  9,  11),  find  the  true  length  of 
cd  and  the  angle  it  makes  with  T'. 

3.  Given  e  (32,  3,  10),  /  (29,  10,  13),  and  g  (24,  5,  18),  find  the 
true  shape  of  the  triangle  efg  by  finding  the  true  lengths  of  its  sides. 

4.  Given  h  (19,  8,  3)  and  i  (11,  11,  5),  find  the  true  length  of  hi 
and  the  angles  it  makes  with  V  and  H.  (Two  operations.)  Compare 
results. 

5.  Given  i  (9,  2,  14)  and  k  (3,  11,  5),  find  the  true  length  oi  jk 
and  the  angle  it  makes  with  P. 

29.  To  Draw  the  Projections  of  a  Line  of  Given  Length, 
Which  Makes  Given  Angles  with  H  and  V.  This  problem  is  the 
converse  of  that  stated  in  §§  24-26.  In  the  former  case,  the 
projections  are  given  and  a  true  length  and  two  true  angles 
are  required.  In  the  present  case,  the  length  and  angles  are 
given  and  the  projections  are  required.  It  is  important  to  study 
this  problem,  since  the  method  can  be  used  in  solving  many 
other  problems  whose  converse  solutions  are  well  known. 

The  method  proposed  is  as  follows.  (1)  Carry  through  a 
solution  of  the  converse  problem  (in  this  case  that  of  §  24), 
without  reference  to  the  quantities  involved,  i.e.  in  the  present 
case,  assume  a  line  at  random,  without  thought  as  to  its  actual 
length  or  inclination.  (2)  Trace  backwards  the  steps  leading 
to  this  solution,  in  the  reverse  order.  Thus,  the  method  and 
order  of  procedure  for  the  required  case  may  be  established. 


34 


DESCRIPTIVE  GEOMETRY 


[III,  §  29 


For  example,  let  it  be  required  to  find  the  projections  of  a 
line  1"  long,  which  makes  an  angle  of  45°  with  H,  and  an  angle 
of  30°  with  V.  Let  the  solution  be  based  on  the  **  first  method  " 
for  finding  true  length  (§  24).  The  operation  consists  in  swing- 
ing the  line  into  parallelism  with  V  (let  us  say),  then  projecting 
it  on  V,  and  thereby  determining  its  true  length  and  slope. 
Applying  this  method  in  the  reverse  sense  (Fig.  30  a),  let  us  draw 
the  projections  of  a  line  1''  long,  making  an  angle  of  45°  with 
H,  and  parallel  to  V,  as  ab'.     If  this  line  is  now  swung  away 


ih  H 


av 


^ 


ib"^ 


^o- 


Fig.  30  a 


Fig.  30  6 


J'h 


-■^ 


from  F,  as  in  the  first  method,  b'"  will  move  toward  z"-',  while 
b'^  moves  in  a  circular  arc  toward  z^.  Meanwhile  the  line 
itself  is  making  constantly  changing  angles  with  V,  some  one 
of  which  is  the  required  angle.  Therefore  the  required  V  pro- 
jection of  b  will  be  somewhere  on  the  line  b'''z%  and  the  required 
H  projection  will  be  somewhere  on  the  arc  b"'z''.  Again,  turn- 
ing to  Fig.  sob,  draw  the  projections  of  a  line  (ab")  1"  long, 
making  an  angle  of  30°  with  V,  and  parallel  to  H.  When  this 
line  is  swung  around  away  from  H,  the  end  marked  b"  will 
trace  out  the  line  b'^'x^  in  horizontal  projection,  and  the  arc 
b^'^x"  in  vertical  projection.      The  required  projections  of  6" 


Ill,  §  29] 


PROJECTION  OF  LINES 


35 


will  be  somewhere  on  the  line  and  the  arc  just  mentioned.  It 
is  not  possible  to  say,  either  in  Fig.  30  a  or  in  30  b,  just  where, 
on  the  lines  or  arcs,  the  required  projections  will  be  found. 
But  by  combining  the  two 
drawings,  as  shown  in  Fig. 
30  c,  the  arc  b'h^  and  the 
line  b'^^x^  will  intersect. 
Since  the  required  H  pro- 
jection of  the  swinging  end 
of  the  line  must  be  on  both 
the  line  and  the  arc,  it  must 
lie  at  the  point  of  intersec- 
tion, b^.  Likewise  the  V 
projection  is  found  to  fall 
at  b\  Note  that  if  the 
construction  is  accurately 
drawn,  the  line  6^6^  will  be 
perpendicular  to  HA.  This  fact  may  be  used  as  a  test  of  accu- 
racy.    More  than  one  solution  is  possible  for  the  given  data. 


Fig.  30  c 


EXERCISE   SHEET   X 

Take  P  at  the  right  border  line.     No  P  projections  are  required. 

1.  The  point  a  (72,  6,  4)  is  given.  The  Une  ab  is  10  units  long  and 
slopes  upward  and  forward  from  a,  making  an  angle  of  30°  with  H  and 
an  angle  of  45°  with  V.     Find  its  projections.     (Two  solutions.) 

2.  The  point  c  (48,  4,  8)  is  given.  The  line  cd,  13  units  long,  slopes 
upward,  forward,  and  away  from  P,  making  an  angle  of  45°  with  H 
and  an  angle  of  15°  with  T^.     Find  its  projections.     (One  solution.) 

3.  The  point  c  (38,  16,  18)  is  given.  The  line  ef  is  8  units  long  and 
makes  an  angle  of  15°  with  H  and  an  angle  of  60°  with  V.  Find  its 
projections.     (Four  solutions.) 

4.  The  point  g  (27,  5,  6)  is  given.  The  line  gh  slopes  upward  and 
forward,  making  an  angle  of  60°  with  H  and  an  angle  of  30°  with  F. 
Draw  the  projections. 

5.  The  point  i  (11,  5,  6)  is  given.  The  hne  ij  is  10  units  long  and 
makes  an  angle  of  60°  with  H  and  45°  with  V.     Explain  the  result. 


36 


DESCRIPTIVE  GEOMETRY 


[III,  §  30 


The  given    distance 


30.   To  Measure  a  Given  Distance  from  a  Given  Point  Along 
a  Given  Line.     In  Fig.  31,  let  the  line  az  be  given  of  indefinite 

length,  and   let  it  be  re- 
.^v    ^  quired  to  mark  off  on  that 

line   a   given  distance,  ac, 
from  a. 

Select  any  other  point 
on  the  line,  as  b.  Find 
the  true  length  of  a6  (§  24), 
as  shown  by  a!"})'^.  On  this 
line  measure  off  the  re- 
quired distance  from  a^  (as 
al'c'^).  Now  rotate  the  line 
back  to  its  first  position. 
The  projections  of  point  c 
will  move  back  along  paths 
parallel  to  those  traced  by 
h^  and  6^'  in  the  first  rota- 
tion. Thus  the  projections 
c^  and  c^  are  determined. 
The  point  c  in  space  is  on  ah,  at  the  required  distance  from  a. 


EXERCISE   SHEET   XI 

Take  P  at  the  right  border  line.     No  P  projections  are  required. 

1.  Given  a  (75,  6,  4)  and  6  (66,  12,  8),  locate  c  on  ah  \"  from  a. 

2.  Given  d  (60,  10,  6)  and  e  (50,  5,  13),  locate  /  on  <h  |"  from  e. 
Find  /  in  two  different  ways,  and  check  results. 

3.  The  hne  gh  slopes  upward  and  toward  V  and  P  from  g  (43,  8,  13), 
making  an  angle  of  30°  with  H  and  an  angle  of  45°  with  T^  Locate  i 
on  gh  7  units  from  h. 

4.  Bisect  the  line  joining  the  points  i  (19,  5,  11)  and  I  (27,  10,  5) 
and  mark  the  rhiddle  point  k.  Compare  the  lengths  y"^•"  and  ^^'",  and 
j^k^  and  V'k^. 

5.  Given  p  (15,  9,  8)  and  m  (4,  14,  14),  divide  pni  into  three  equal 
parts,  marking  the  points  of  division  n  and  o.  State  the  ratios  between 
the  segments  of  the  projections  and  those  of  the  true  length. 


Ill,  §  31] 


PROJECTION   OF  LINES 


37 


31.  Traces  of  a  Line.  Every  normal-case  line,  if  sufficiently 
extended,  meets  the  planes  of  projection  in  points  which  are 
called  the  traces,  or  piercing  points,  of  the  line.  (See  Fig.  32  a.) 
They  are  especially  significant  and  useful.  Both  projections  of 
a  trace  lie  on  the  projections  of  the  line,  and  one  of  them  is 
always  in  HA,  since  a  trace  is  always  in  a  plane  of  projection. 


Fig.  32  6 


®  indicates'  a   Trace  o 

vf  =  vertical    trace  . 

ht  =  horizontd!    trace 

vf  -  horizontal    projection  of  veriical     trace. 

ht^=  vertical     projection  of    horizontal  trace. 

Fig.  32  a 


Figure  32  b  shows  in  projection  the  same  line  as  was  used  in 
32  a.  Let  it  be  required  to  find  its  traces  As  the  line  is  ex- 
tended from  a  toward  V  (Fig.  32  a),  the  H  projection  is  extended 
from  a^  toward  HA.  \Mien  the  line  touches  V  (at  vt),  the  pro- 
jection intersects  HA  (at  vt^).  The  equivalent  operation  in 
Fig.  32  6  consists  in  extending  b^a^  to  HA,  giving  vt^  as  the  H 
projection  of  the  V  trace.  The  V  trace  is  on  the  V  projection 
of  the  Hne  directly  above  vt^.  The  H  trace  {ht)  is  found  in  a 
similar  manner  by  extending  a^'b''  to  HA  and  dropping  a  per- 
pendicular to  a^b^.  When  three  planes  of  projection  are  used 
there  will  of  course  be  three  traces.    This  case  is  treated  in  §  44. 


38 


DESCRIPTIVE   GEOMETRY 


[III,  §  32 


32.  Traces  of  a  Line  Parallel  to  P.     If  the  given  line  is  parallel 

to  P,  its  traces  cannot  be  found  in  the  manner  described  above. 

But  if  a  profile  projection  is  constructed,  the  traces  are  easily 

found. 

EXERCISE    SHEET    XII 

Take  P  at  the  left,  16  from  the  border  line  and  turned  away  from 
the  objects,  into  V.     Draw  the  P  projections  only  when  necessary. 

1.  Given  a  (4,  13,  3)  and  h  (10,  2,  6),  find  the  H  and  V  traces  of  ah. 

2.  Given  c  (15,  9,  5)  and  d  (24,  9,  5),  find  the  trace  of  cd. 

3.  Given  e  (28,  4,  3)  and/  (37,  10,  4),  find  the  H  trace  of  ef. 

4.  Given  g  (40,  3,  10)  and  h  (40,  16,  3),  find  the  H  and  V  traces  of  gh, 

5.  Given  i  (48,  2,  13)  andj  (58,  8,  4),  find  the  H  and  V  traces  of  ij, 
and  locate  /:  on  ij  2"  from  the  H  trace. 

33.  Intersecting  Lines.  Two  lines  which  Intersect  in  space 
have  a  point  in  common.  The  projections  of  this  point  must 
lie  on  the  projections  of  each  of  the  lines.     Also  (see  3,  Art.  13) 


A       H 


Fig.  33  a 


Fig.  33  6 


the  H  and  V  projections  of  the  point  of  intersection  must  lie 
on  the  same  perpendicular  to  HA. 

Figure  33  a  shows  the  projections  of  two  lines  ah  and  cd,  which 
intersect  at  e.  In  Fig.  33  h  the  lines  fg  and  l:j,  as  seen  on  F, 
appear  to  intersect  at  m.  If,  however,  a  perpendicular  is 
dropped  from  m""  it  will  be  seen  that  the    V  projection,  iii^j 


Ill,  §  34]  PROJECTION  OF  LINES  39 

corresponds  to  two  //  projections,  m^  and  mi".  This  means  that 
though  fg  and  kj  appear  (on  V)  to  intersect  at  m,  in  reality  jg 
passes  well  in  front  of  kj  at  that  height. 

Similarly  the  lines  appear  (on  H)  to  intersect  at  n,  while  in 
reality  kj  passes  well  below  fg.  By  raising  kj  or  by  drawing 
it  away  from  V  or  both,  the  lines  can  be  made  to  intersect. 
In  this  event  the  points  m  and  n  will  come  together. 

EXERCISE    SHEET    XIII 

Take  P  at  the  right,  16  from  the  border  hne,  and  turned  away  from 
the  objects,  into  T'.     Use  P  only  when  necessary. 

1.  Does  the  line  joining  a  (GO,  10,  6)  and  h  (51,  3,  5)  intersect  the 
line  joining  c  (60,  4,  3)  and  d  (51,  8,  9)? 

2.  In  the  space  between  47  and  37  from  P  draw  the  projections 
of  two  lines  fg  and  hi  which  intersect  at  a  point  e,  43  from  P.  From 
/,  jg  slopes  away  from  H  and  V  and  toward  P ;  hi  slopes  from  h  towards 
H,  y,  and  P. 

3.  Given  j  (33,  6,  4)  and  k  (25,  20,  10),  locate  the  point  I  on  jk 
12  units  from  J,  and  through  /  draw  a  hne  mn  parallel  to  HA. 

4.  Given  o  (16,  15,  8)  and  p  (4,  7,  10),  through  g,  the  middle  point 
of  op  draw  rs  15  units  long,  perpendicular  to  F  and  one  end  resting 
on  F.  Draw  an}'  other  line  tu  intersecting  both  op  and  rs.  Draw  all 
three  projections,  and  check  for  accurac}'. 

34.  Parallel  Lines.  Any  two  lines  that  are  parallel  in  space 
will  be  projected  on  H  by  parallel  projecting  planes.  (Why?) 
The  lines  in  which  these  planes  are  cut  by  H  are  the  H  pro- 
jections of  the  given  lines ;  hence  they  wall  be  parallel.  The 
same  reasoning  will  apply  with  respect  to  the  V  and  P  pro- 
jections. Therefore,  we  may  say  that  the  corresponding 
projections  of  parallel  lines  are  parallel. 

EXERCISE    SHEET    XIV 
Take  P  at  the  right  border  Une,  turned  toward  the  objects,  into  T'. 

1.  Given  a  (76,  5,  8)  and  h  (65,  10,  4),  draw  all  three  projections 
of  ah  and  also  of  cd,  parallel  to  ah,  but  farther  from  both  H  and  F  than 
is  ah. 


40 


DESCRIPTIVE   CxEOMETRY 


[III,  §  34 


2.  Given  e  (60,  12,  1)  andf  (49,  3,  12),  through  g,  on  ef,  1"  from/, 
draw  /ii  parallel  to  ah. 

3.  Through  /  (39,  6,  0),  draw  mn  parallel  to  the  line  joining  j  (44, 
16,  14)  and^  (35,  7,  14). 

4.  Draw  op  parallel  to  ef  and  oq  parallel  to  ab  through  o  (23,  12,  9). 
Find  the  true  size  of  the  angle  poq. 

[Note.     Draw  a  third  line  intersecting  op  and  oq.     Find  the  true 
length  of  the  three  lines  and  construct  the  triangle.] 

35.    Angle  between  Two  Lines.     In  general  the  angle  between 
two  lines  is  not  shown  in  its  true  size  bv  the  ande  between  the 


Fig.  34 

projections  of  the  lines.  Let  the  student  prove  this  proposition 
by  the  use  of  Fig.  34. 

An  exception  to  the  preceding  statement  occurs  when  both 
the  lines  are  parallel  to  the  plane  on  which  the  projection  is 
made.  A  method  for  determining  the  true  angle  between  any 
two  lines  is  given  in  §  94. 

36.  Perpendicular  Lines.  In  the  case  of  perpendicular  lines 
the  projections  are  perpendicular  only  when  either  or  both  of 
the  lines  are  parallel  to  the  plane  of  projection.  To  prove  this, 
begin  by  assuming  that  the  projections  are  perpendicular,  and 
that  one  of  the  original  lines  is  not  parallel  to  the  plane  of  pro- 
jection, and  show  that  the  second  one  of  the  original  lines  must 
be  parallel  to  the  plane  of  projection. 


Ill,  §  37]  PROJECTION   OF   LINES  41 

37.    Recapitulation  of  Chapter  III. 

1.  The  projections  of  a  /me  are  determined  by  the  projections 
of  any  two  of  its  points. 

2.  Any  two  lines,  one  of  which  is  drawn  on  H  and  the  other  on  V, 
will  represent  the  projections  of  some  line  in  space. 

3.  The  lengths  and  the  slopes  and  the  angles  between  lines  are 
not,  in  general,  shown  in  their  true  size  by  the  projections  of  the 
lines. 

4.  Special  cases  form  exceptions  to  3,  above. 

5.  By  artificially  reducing  normal  cases  to  special  cases,  true 
lengths  and  slopes  may  be  detcrnmied. 

6.  The  traces  of  a  line  are  points  of  special  significance. 

7.  Apparent  intersections  are  not  necessarily  real  ones. 

8.  Projections  of  parallel  lines  are  parallel. 

9.  Projections  of  perpendicular  lines  are  not,  in  general, 
perpendicular. 

10.    There  are  exceptions  to  9  above. 

[Note.     At  this  point  the  student  may  take  up  with  profit  §§  179- 
182  of  Chapter  IX  and  §§  202-210  of  Chapter  XII.] 


CHAPTER  IV 


PROJECTION    OF   POINTS   AND   LINES   IN   THE   SECOND, 
THIRD,    AND    FOURTH    QUADRANTS 

38.  Introduction.  It  is  often  necessan-  to  follow  a  line  be- 
yond the  limits  of  the  solid  angle  which  we  have  considered  thus 
far  as  Hmiting  the  field  of  operations.     Such  an  extension  is 


Line    ab     ihowinq   its   V   trace 


Plan  of   line   ab 
a" 


H 


Elevation  of    line    ab 
(b) 


Fig.  35 


required  both  in  abstract  problems  and  in  their  practical  appli- 
cations, such  as  casting  shadows. 

For  example,  suppose  that  it  is  required  to  find  the  //  trace 
of  ah,  Fig.  35  a.  If  the  line  is  extended  beyond  h,  it  is  seen  to 
intersect  V  at  vt.  From  the  projections,  it  can  be  seen  that  as 
ab  approaches  V,  it  also  approaches  H.  But  it  intersects  V 
before  it  reaches  //  ;  hence  if  it  intersects  H  at  all,  it  must  do  so 
at  some  point  behind  V. 

42 


IV,  §  38] 


VARIOUS  QUADRANTS 


43 


To  make  this  clearer,  the  plan  of  ah  is  drawn  in  Fig.  35  c,  sep- 
arately from  the  elevation  which  is  illustrated  in  Fig.  35  h. 
Now  let  both  projections  be  extended  indefinitely  beyond  h 
as  shown.  The  line  will  pass  through  T^  at  the  point  vt  as  pre- 
viously determined,  and  will  touch  H  at  the  point  hi,  which  is 
determined  by  the  intersection  of  a'V'  with  HA. 

Figure  35  d  shows  the  operations  which  were  performed  sep- 
arately in  Figs.  35  h  and  35  c,  carried  out  on  a  single  HA,  i.e. 


the  two  figures  35  h  and  35  c  are  drawn  together  until  the  two  HA 
lines  coincide.  In  Fig.  36  the  same  line  with  its  two  traces  is 
shown  in  isometric  projection. 

In  order  to  determine  the  H  trace  of  ah,  it  was  necessary  to 
extend  the  H  plane  back  beyond  T"  so  as  to  create  a  new  solid 
angle  or  quadrant  previously  unnoticed. 

While  we  are  thus  extending  the  scope  of  our  operations, 
we  shall  reach  the  possible  limit  by  also  extending  T"  below  //, 
as  shown  by  the  dotted  lines  in  Fig.  36.  If  now  the  planes  are 
considered  as  indefinite  in  extent,  every  point  in  space  may 
be  dealt  with  freely,  since  all  space  is  contained  within  one  of 
the  four  quadrants  thus  formed. 


44 


DESCRIPTI\^  GEOMETRY 


[IV,  §  38 


These  four  quadrants  are  numbered,  for  easy  reference,  as 
shown  in  Fig.  37.     In  most  problems  it  is  possible  to  choose  the 

planes  of  projection  so  that  all 
the  work  falls  within  the  first 
quadrant.  Most  of  the  exercises 
in  the  following  chapters  are  so 
chosen.  However,  as  noted 
above,  it  is  sometimes  necessary 
to  deal  with  parts  of  lines  or  with 
points  in  the  second,  third,  and 
fourth  quadrants. 

39.    Projecting   a   Point.     The 

operation  of  projecting  a  point  is 

the  same  in  all  the  quadrants.     The  projecting  lines  in  every 

case  pass  through  the  point  and  are  perpendicular  to  the  plane 

of  projection.     (See  Fig.  38.)     This  means,  of  course,  that  the 


Fig.  37 


Fig.  38 


IV,  §  40] 


VARIOUS   QUADRANTS 


45 


H  projecting  lines  for  points  in  quadrants  I  and  II  are  drawn 
downward,  while  for  points  in  quadrants  III  and  IV  they  are 
drawn  upward.  Similarly  in  quadrants  I  and  IV,  the  V 
projecting  lines  are  drawn  away  from  the  observer  while  in  quad- 
rants II  and  III  they  are  drawn  toward  the  observer. 


fC 


H- 


b' 


Fio.  39 


■A 


Fig.  40 


40.  Opening  the  Planes.  The  manner  of  opening  the  planes 
previously  established  for  the  first  quadrant  is  retained,  i.e. 
the  part  of  V  which  is  above  H  is  rotated  backward  into  H. 
At  the  same  time  that  part  of  V  which  lies  below  H  rotates  for- 
ward and  upward  into  H.  In  other  words,  the  entire  V  plane 
is  rotated  on  HA  as  an  axis,  the  upper  part  following  the  rule  for 
the  first  quadrant  (Fig.  40). 

The  effect  of  this  rotation  is  to  "  open  "  the  first  and  third  and 
to  "  close  "  the  second  and  fourth  quadrants.^  When  the  planes 
are  opened  as  described  above,  it  will  be  seen  that  the  part  of 
the  resulting  sheet  which  lies  above  HA  contains  the  upper  part 
of  V  and  the  rear  part  of  H,  while  below  HA  are  found  the  front 
part  of  H  and  the  lower  part  of  V. 

^  Sometimes,  particularly  in  blackboard  work,  it  may  be  more  natural 
to  think  of  the  forward  part  of  H  as  being  rotated  downward  and  backward 
into  V.  The  net  result  in  either  case  is  the  same,  viz.,  to  open  the  first 
and  third  and  to  close  the  second  and  fourth  quadrants. 


46 


DESCRIPTIVE  GEOMETRY 


[IV,  §  40 


The  student  should  now  make  himself  a  pair  of  planes.  (See 
Fig.  41.)  These  may  be  used  to  assist  in  visualizing  the  prob- 
lems and  in  following  the  text. 


Cui 


9 


Open 


Cut    two    pieces  of  card- 
boord    as   shown    and  dip 
fogeiher    ihrough     sIoId. 


Fig.  41 

41.  The  Typical  Projections.  Of  course,  the  projections 
that  are  made  on  the  planes  previous  to  the  opening  or  flattening 
out  process  move  with  the  planes.  The  opening  of  the  first 
and  third  quadrants  moves  the  //  and  T^  projections  to  opposite 
sides  of  HA.  In  the  case  of  the  third  quadrant,  the  V  projection 
lies  below  HA  instead  of  above,  as  in  the  case  of  first  quadrant 
points.  On  the  other  hand,  the  closing  of  the  second  and  fourth 
quadrants  moves  both  projections  to  the  same  side  of  HA,  but 
in  one  case  (fourth)  both  projections  fall  below  HA,  whereas  in 
the  other  case  (second)  both  fall  above.  To  assist  In  visualizing 
these,  relations,  refer  to  Figs.  38  and  39,  where  both  isometric 
and  orthographic  projections  of  the  same  points  are  shown 
together. 

Figure  42  shows  the  projections  of  four  points,  a,  b,  c,  and  d, 
located  at  the  same  distances  from  H  and  V,  but  each  in  a  dif- 
ferent quadrant,  as  shown  by  the  Roman  numeral.  These 
projections  illustrate  the  relations  pointed  out  above. 


IV,  §  41] 


VARIOUS   QUADRANTS 


47 


In  solving  the  first  few  problems  dealing  with  projection  in 
all  quadrants,  the  student  may  be  helped  by  the  pairing  of  the 
quadrants  suggested  above,  first  with  third  and  second  with 
fourth.  Each  quadrant  has  a  similar  and  an  opposite  charac- 
teristic to  those  of  its  pair-mate.  The  first  and  third  quadrants 
are  similar  in  that  the  two  main  projections  ( //  and  V)  are  sep- 
arated by  HA,  but  are  opposite  in  that  the  V  (or  H)  projection 
is  above  in  one  case  and  below  in  the  other. 

The  second  and  fourth  quadrants  are  similar  in  that  the  two 
main  projections  are  on  the  same  side  of  HA,  but  are  opposite 


I 


m      11 


H 


b^ 


Fig.  42 


in  that  in  one  case  the  projections  are  above  while  in  the  other 
case  they  are  below  HA. 

Throughout  these  relations  of  similarity  and  oppositeness, 
however,  there  are  certain  fundamental  relations  which  hold 
good  for  any  quadrant.  Thus,  the  nine  principles  of  §  13  are 
applicable  in  any  quadrant. 

[Note.  In  making  use  of  this  pairing  of  the  quadrants  and  the 
word  descriptions  of  the  typical  projections,  the  student  should  remem- 
ber that  there  is  little  virtue  in  memorizing  these  characteristic  relations 
as  set  formulas.  Unless  the  student  clearly  realizes  the  fact  that  these 
relations  are  the  necessary  result  of  the  method  used  in  projecting  the 
points  and  in  opening  the  planes,  and  unless  the  whole  process  is  clearly 
visualized,  no  satisfactory  progress  can  be  anticipated.] 


48 


DESCRIPTIVE   GEOMETRY 


[IV,  §  42 


42.  Opening  the  Profile  Plane.  The  profile  plane  is  used 
for  projections  in  all  four  quadrants  in  the  manner  shown  in 
Fig.  43.  There  are  four  possible  methods  of  opening  the  profile 
plane  in  this  case,  just  as  there  are 
four  possible  methods  when  only 
one  quadrant  is  used  (§  11). 

Figures  44  and  45  show  the  open- 
ing of    P  to  the   left  into   V. 
Figure  46  shows  the  projec- 
tions  of   the   same 
points   as  are 
shown     in 
43,  when  P 
is  opened 
in       the 
manner  shown 
in  Figs.  44  and  45. 

It  will  be  much 
easier    to    follow 
this  operation  in  pro- 
jection  if    the   signifi- 
cance   of    the    dotted 
quarter  circles  is  fully 

understood.  Referring  to  Fig.  46,  let  the  student  think  of  him- 
self as  viewing  the  planes  of  projection  from  above.  Now  HP' 
represents  the  top  view  of  F,  and  VV  the  top  view  of  P.    From 


Fig.  43 


Fig.  44 


Fig.  45 


IV,  §  42] 


VARIOUS  QUADRANTS 


49 


this  view  point,  any  point  in  space,  as  a,  ^Yill  be  directly  above 
its  H  projection,  a^.  Likemse  the  V  projection  of  the  point 
will  be  in  HP\  as  r,  and  the  P  projection  will  be  in  VV\  as  vk 

Xow  let  the  operation  of  Fig.  44  be  performed.  The  front  part 
of  P,  carrying  the  point  m  with  it,  will  move  toward  left  through 
an  arc  of  90°,  stopping  at  ?u  This  shows  the  significance  of  the 
arc  mn  :  it  represents  the  path  of  a^  during  the  first  operation  of 
opening  the  P  plane. 

The  operation  shown  in  Fig.  45  starts  with  a^  in  the  position 
n  (as  seen  from  above),  considerably  above  H.     As  V  and  P 


-\/-- 


H 


^ 


H      ^ 


■V^-.tb" 
I 


m 


P' 


*c 


Fig.  46 

are  rotated  into  H,  the  point  a  describes  another  quarter  circle, 
whose  plane  is  perpendicular  to  HA.     It  is  shown  by  the  Hne 

In  the  case  of  the  point  d  (fourth  quadrant)  the  line  od^ 
represents  tlbe  arc  of  the  second  operation  (as  does  naP  above), 
but  this  line  is  drawn  in  the  opposite  direction  from  HA,  since 
it  represents  the  motion  of  the  lower  half  of  the  T^  and  P  planes. 

The  student  should  now  follow  out  the  significance  of  the 
arcs  for  points  b  and  c;  and  he  should  also  work  out  typical 
projections  for  the  cases  where  P  is  opened  to  the  right  into  V, 
and  both  to  the  right  and  to  the  left  into  H. 


50 


DESCRIPTIVE  GEOMETRY 


[IV,  §  42 


EXERCISE    SHEET    XV 

Take  P  at  the  left,  16  from  the  border  line,  and  turned  away  from 
the  objects,  into  T^.^ 

Draw  all  three  projections  of  the  following  points : 

a  (16,  16,  9)  e  (II,  32,  7,  13) 

h  (II,  20,  13,  5)  /  (III,  36,  17,  3) 

c  (III,  24,  5,  10)  g  (40,  23,  4) 

d  (IV,  28,  11,  7)  h  (IV,  44,  7,  13) 

2,  Given  the  point  i  (48,  8,  6),  move  i  parallel  to  HA  till  it  is  4 
farther  from  P.  From  this  position  move  it  straight  back  12,  and  then 
down  30. 

3.  Given  the  point  j  (IV,  60,  13,  13),  move  j  straight  toward  h  of 
Ex.  1,  until  it  is  56  from  P. 

43.  Projection  of  Lines.  The  projection  of  lines  passing 
through  more  than  one  quadrant  will  give  no  trouble  if  it  is 

remembered    that   two 
'^  points     always     deter- 

mine a  line,  and  that 
the  projection  of  a  line 
always  passes  through 
the  corresponding  pro- 
jections of  the  points. 

Figure  47  shows  the 
line  ah  passing  from  a 
in  the  first  quadrant  to 
h  in  the  third.  In  Fig. 
48  a  the  projections  of 
the  points  /i  and  h  are 
shown.  Lines  drawn 
through  a!"  and  h^  and  also  through  a"  and  6^  are  the  projec- 
tions of  the  lines  ah.  The  traces  of  ah  are  at  vt  and  ht  (§  31). 
From  Fig.  47,  it  will  be  seen  that  ah  passes  from  the  first  quad- 

*  In  describing  the  movement  of  P,  the  description  will  be  understood  to 
apply  to  the  movement  of  that  part  of  P  which  lies  in  the  first  quadrant,  as 
heretofore. 


Fig.  47 


IV,  §  44] 


VARIOUS   QUADRANTS 


51 


rant,  through   V,  and   hence  into  the  second   quadrant;    and 
from  there  through  H,  and  hence  into  the  third  quadrant. 

If  now  we  turn  to  Fig.  48  a,  it  will  be  seen  that  all  points 
between  vt  and  ht  do  actually  correspond  to  the  typical  case  for 
points  in  the  second  quadrant,  i.e.  both  projections  are  above 
HA.  Similarly,  all  points  between  ht  and  b  are  seen  to  be  points 
in  the  third  quadrant  and  all  points  between  vt  and  a  are  points 
in  the  first  quadrant, 
since  these  points  corre- 
spond to  the  typical  cases 
for  these  quadrants. 

Turning  again  to  Fig. 
47,  it  will  be  seen  that  if 
the  point  b  is  gradually 
lowered  vertically,  the 
traces  ht  and  vt  gradually 
approach  HA,  and  one 
another.  Soon  the  traces 
will  coincide,  i.e.  the  line 
will   pass   through   HA. 

If  the  lowering  of  b  is  -pic.  4S  h 

continued    beyond    that 

point,  the  line  will  pass  from  the  first  quadrant  tlirough  H  into 
the  fourth,  and  thence  through  T^  into  the  third.  The  H  trace, 
ht,  has  crossed  to  the  right  of  vt.  This  case  is  shown  in 
Fig.  48  b. 

The  student  should  analyze  this  latter  case  as  we  did  above 
for  Fig.  48  a.  He  should  then  try  to  draw  a  line  passing  from 
the  second,  through  the  first  and  thence  into  the  fourth 
quadrant;  and  as  many  other  cases  as  may  suggest  them- 
selves. 

44.  Profile  Trace  of  a  Line.  In  §  31  only  two  traces  of  a  line 
were  mentioned.  Any  normal-case  line  will  intersect  all  three 
planes  of  projection,  and  hence  it  will  have  three  traces. 


52 


DESCRIPTIVE   GEOMETRY 


[IV,  §  44 


In  Fig.  49,  the  line  ah  pierces  V  at  vt,  H  at  ht,  and  P  at  pt. 

Follow  the  line  out 

a'  Q''  first  on  V,  then  on 

// ;  and  visualize 
it.  Remembering 
that  pt^  and  pt^  are 
the  H  and  V  pro- 
jections of  the  P 
trace,  construct  the 
P  projection  of 
the  line.  Then  the 
position  of  the  P 
trace  is  apparent. 
EXERCISE    SHEET    XVI 

Take  P  at  the  right,  16  from  the  border  line,  and  turned  away  from 
the  objects,  into  H.     Draw  all  three  projections  of  the  following  lines  : 

1.  Given  a  (60,  15,  2)  and  b  (52,  3,  10),  find  the  H  and  V  traces 
of  ab. 

2.  Given  c  (II,  48,  6,  4)  and  d  (IV,  40,  3,  9),  find  the  traces  of  cd, 
and  mark  the  quadrants  traversed  by  the  line. 

3.  Given  e  (37,  7,  12)  and/  (III,  37,  9,  8),  find  the  traces  of  ef. 

4.  Given  g  (III,  30,  4,  1)  and  h  (III,  20,  14,  6),  through  a  point  i 
on  gh,  10  imits  from  g,  draw  a  line  parallel  to  ab.  Find  its  traces,  and 
mark  them  r  and  s.     Draw  the  profile  projection  of  rs  only. 

EXERCISE    SHEET    XVII 

Take  P  at  the  left,  16  from  the  border  line,  and  turned  toward  the 
objects,  into  //. 

1.  Given  a  (4,  11,  5)  and  b  (III,  10,  2,  4),  draw  the  three  projections 
of  ab  and  find  all  three  traces  of  ab. 

2.  Through  c  (II,  15,  16,  20),  draw  cd  10  units  long,  making  angles 
of  45°  and  30°  with  H  and  V,  respectively.  From  c,  cd  slopes  away 
from  P  and  H  and  toward  V. 

3.  Given  e  (27,  3,  7)  and/  (III,  36,  5,  6),  find  the  true  length  of  ef 
by  the  second  method. 

4.  Through  g  (IV,  50,  5,  12),  draw  gh  parallel  to  cd,  h  being  in  I 
at  a  distance  45  from  P.  Find  the  //  and  V  traces  of  gh,  the  true 
length  of  gh,  and  the  angles  it  makes  with  //  and  V.     Check  with  cd. 


CHAPTER  V 


PLANE  FIGURES   AND    SOLIDS 

45.  Projection  of  Plane  Figures.  Limiting  Conditions. 
Any  plane  figure  which  is  parallel  to  a  plane  of  projection  will 
be  projected  on  that  plane  in  its  true  size  and  shape.  The 
projecting  planes  of  the  lines  which  form  the  sides  of  such 
a  figure  form 
a  projecting 
prism.  (See 
Fig.  50  a.) 
Both  the  fig- 
ure itself  and 
its  projection 
are  right  sec- 
tions of  the 
projecting 
prism,  i.e.  sec- 
tions perpen- 
dicular to  the 
prism's  axis. 

If   now   the 

figure  is  imagined  to  be  slowly  tilted  toward  the  plane  of  pro- 
jection (Fig.  50  6),  the  projecting  prism  will  contract  in  thick- 
ness. The  figure  in  space  is  now  an  oblique  section  of  the 
projecting  prism,  while  its  projection,  which  remains  a  right 
section,  is  smaller  than  the  figure  itself  by  an  amount  that  de- 
pends upon  the  angle  of  inclination.  The  limiting  case  will 
occur  when  the  figure  is  perpendicular  to  the  plane  of  pro- 
jection, the  projecting  prism  having  become  a  plane  and  the 
projection  having  become  a  line. 

53 


54 


DESCRIPTIVE   GEO:\IETRY 


[V,  §  46 


46.  Special  Case.^  An  adaptation  of  the  principles  of  §  27 
enables  us  to  construct  the  projections  of  any  plane  figure,  one 
side  of  which  is  parallel  to  a  plane  of  projection. 

I.  Let  it  be  required  to  draw  the  projections  of  a  regular 
pentagon  one  side  of  which  is  parallel  to   H  and  V,  and  the 


plane  of  which  is  inchned  at  an  angle  of  30°  to  H  (Fig.  51). 
In  this  case,  the  plane  of  the  figure  is  perpendicular  to  P.  Hence 
the  P  projection  will  be  a  line  inclined  at  an  angle  of  30°  to  PA. 
Let  us  start  the  drawing  on  P,  using  a  P  plane  opened  to  the 
right  into  H.  Any  line  of  indefinite  length,  such  as  a^h^,  drawn 
on  P  and  making  an  angle  of  30°  with  PA,  may  represent  this 
projection.  It  is  now  necessary  to  locate  on  this  line  points  which 
will  indicate  the  vertices  of  the  pentagon.  In  any  convenient 
position  draw  a  pentagon  of  the  required  size,  as  c  —  g.  Measure 
off  the  distances  hj  and  jf  on  a''%^.  In  Fig.  51  the  pentagon 
is  drawn  with  the  side  dc  perpendicular  to  a^h'^  and  with  the 
median  line  hf  parallel  to  (O'h^.  It  is  not  necessary  to  draw  the 
polygon  in  this  position,  however.  It  may  be  drawn  anywhere 
^  For  the  general  case,  see  §  96. 


V,  §  46] 


PLANE  FIGURES  AND  SOLIDS 


55 


on  the  sheet,  and  the  distances  hj  and  jf  may  be  transferred  to 
aPbP  by  means  of  dividers. 

Next  locate  the  median  Hne  ab  on  plan  and  elevation  at  any 
convenient  distance  from  P  {a^h^,  a>'6').  From  the  known 
P  projections  of  h,  j,  and  /,  locate  the  H  and  V  projections  of 
these  points  on  a^b^  and  a^'b^'. 
Now  cb  and  gc  will  be  parallel 
to  H  and  T^ ;  hence  their 
H  and  V  projections  will 
show  their  true  lengths,  and 
will  be  parallel  to  HA. 
Draw  these  lines,  mark  off 
their  end  points,  and  con- 
nect the  points  so  found. 

II.  A  slightly  different 
construction  must  be  used  if 
the  plane  of  the  figure  is  in- 
clined toward  all  three  of 
the  planes  of  projection. 
For  example,  let  it  be  re- 
quired to  draw  the  projections  of  a  square  with  one  side  on  H, 
inclined  at  an  angle  of  45°  to  T',  and  with  the  plane  of  the 
figure  inclined  at  an  angle  of  60°  to  //.     (See  Fig.  52.) 

Start  with  the  projection  on  a  V  plane,  perpendicular  to  the 
plane  of  the  figure  as  indicated  by  the  axis  H'A'.  The  line 
a^-V,  drawn  at  an  angle  of  60°  to  H'A',  and  in  length  equal  to 
a  side  of  the  square,  is  such  a  projection.  One  side  of  the  H 
projection  (aV)  is  known  in  length  and  in  direction ;  hence  that 
side  can  be  drawn.  The  angles  of  the  square  will  project  in 
their  true  size,  by  §  36.  The  H  projection  can  now  be  com- 
pleted by  drawing  a  perpendicular  through  b'',  thus  determining 
b^  and  d^.  Knowing  the  plan,  the  elevation  may  be  drawn  by 
projecting  the  points  upward  and  measuring  the  heights  from 
the  F'  projection,  as  in  §  27. 


.B"d^' 


56 


DESCRIPTIVE  GEOMETRY 


[V,  §  47 


47.  Plane  Solids.  Solids  that  are  determined  by  plane 
surfaces  are  called  plane  solids.  There  are  two  general  types, 
regular  and   irregular.     The   regular   solids   are   those  whose 


Tetrahedron 


Cube 


Octahedron 
Fig.  53 


Dodecahedron 


Icosahedron 


faces  are  regular  polygons,  so  joined  together  that  the  angles 
between  the  planes  are  equal,  and  that  no  one  of  the  planes,  if 
prolonged,  will  cut  into  the  solid.     Besides  the  regular  solids 

^,. ,.^  there  are  prisms,  pyramids, 

and   an  infinite  number  of 
other  irregular  shapes. 

48.  The  Regular  Solids. 
There  exist  in  all  just  five 
regular  solids.  (See  Fig. 
53.)  They  are  named  as 
follows  : 

(1)  The  tetrahedron, 
bounded  by  four  equilateral 

h    triangles. 

(2)  The  hexahedron,  or 
cube,  bounded  by  six 
squares. 

(3)  The  octahedron, 
bounded  by  eight  equi- 
lateral triangles. 

(4)  The      dodecahedron, 
bounded  by  twelve  regular  pentagons. 

(5)  The  icosahedron,  bounded  by  twenty  equilateral  triangles. 

49.    Development.     If  the  surface  of  a  given  solid  can  be 

flattened  out  on  a  plane  without  distortion,  that  surface  is  said 


H                 3'd' 

/K 

zz                 A 

^ 

/  '  \ 
/   •   \ 

b"                 i 

\ii(o^ 

^/" 

d' 

\j/ 

Fig.  54 


V,  §  49] 


PLANE  FIGURES  AND  SOLIDS 


57 


to  be  developable.  The  making  of  collapsible  paper  boxes  illus- 
trates the  point.  A  sphere  could  not  be  made  in  the  same  way. 
This  idea  is  important  in  the  sheet  metal  trade. 


In  Fig.  54,  for  example,  the  plan  and  elevation  of  a  square 
pyramid  are  shown.  To  develop  the  surface,  let  each  of  the 
sloping  faces  be  rotated  outward,  until  they  fall  in  the  plane  of 
the  base.     The  figure  a^e  —  d^f  is  the  required  development. 


Fig.  56 


Fig.  57 


If  the  pyramid  is  placed  with  one  of  its  triangular  faces  on 
the  plane  of  development,  and  then  is  rolled  over  so  that  each 
face  in  turn  is  brought  into  this  plane,  the  development  appears 
as  in  Fig.  55.  Figures  56  and  57  show  the  developments  of  an 
octahedron  and  an  icosahedron,  respectively. 


58 


DESCRIPTIVE  GEOMETRY 


[V,  §  50 


50.  To  Construct  the  Projections  of  a  Tetrahedron.  In 
whatever  position  a  tetrahedron  may  be  placed  with  reference 
to  the  planes  of  projection,  some  of  its  edges  will  be  normal-case 
lines ;  hence  they  will  be  foreshortened  in  projection.  The 
problem  of  drawing  the  projections  of  such  a  figure  is  therefore 
not  simple.  * 

Let  it  be  required  to  draw  the  projections  of  a  regular  tetra- 
hedron 1"  on  each  side  and  resting  on  H.     (See  Fig.  58.)     The 


plan  can  be  drawn  from  considerations  of  symmetry.  The 
elevation  is  started  by  locating  a^'6^  and  c^,  which  are  known  to 
lie  in  HA.  The  determination  of  the  height  of  the  point  o 
presents  some  difficulty,  since  it  is  connected  to  the  points  so 
far  determined  by  lines  that  will  be  foreshortened  in  projection. 
The  principle  of  the  following  solution  is  that  of  §  24.  Let 
ao  be  rotated  about  a,  until  it  is  parallel  to  V.  Its  plan  is  now 
at  a^o'^.  The  V  projection  of  o'  will  lie  on  o'^'z.  Moreover, 
the  V  projection  of  the  rotated  line  will  show  its  true  length. 
Therefore  o'  will  lie  on  an  arc  swung  from  a"-',  with  a  radius 
of  1"  (as  1-2).     The  intersection  of  this  arc  with  o'^'z  will  give 


V,  §  51] 


PLAXE   FIGURES  AND   SOLIDS 


59 


the  point  o''.  Now  if  the  line  is  swung  back  to  its  correct 
position,  o''  will  trace  a  horizontal  line,  and  o''  will  be  on  this 
line  and  directly  above  o^.  Having  determined  o%  connect  it 
with  a'',  c'',  and  b'\  The  projections  of  the  tetrahedron  are  now 
determined.  From  the  H  and  V  projections  the  P  projection 
can  be  determined  if  it  is  required. 

51.   To  Construct  a  Solid  from  Its  Development.     For  the 
purpose    of    illustration    a    dodecahedron   will    be    used.     The 


dT' 


Fig.  59 


general  method  for  this  prol^lem  is  to  work  from  the  known 
converse  solution.  (See  §  29.)  In  this  case  the  development 
of  one  half  of  the  solid  (consisting  of  six  pentagons)  will  be  as 
shown  by  the  dotted  lines  in  Fig.  59.  The  outside  figures  will 
be  rotated  about  the  lines  joining  them  to  the  central  figure  until 
corresponding  points  (as  /  and  g,  k  and  j,  etc.)  fall  together. 
This  operation  will  be  seen  to  be  exactly  the  reverse  of  the 
process  for  making  a  development. 

To  follow  this  operation  in  detail,  think  of  the  development 
in  Fig.  59  as  on  the  H  plane.     Let   the  central  figure  remain 


60 


DESCRIPTIVE   GEOMETRY 


[V,  §  51 


on  H,  as  above,  while  the  figures  surrounding  it  are  rotated  up- 
ward, swinging  on  the  side  which  touches  the  base  as  an  axis. 
Each  of  the  vertices  will  move  in  a  circular  arc  which  is  pro- 
jected on  plan  in  a  straight  line  perpendicular  to  its  axis  of 
revolution,  as  shown  in  the  figure  by  the  dotted  lines  j^2^,  k^2^, 
/1\  fl^,  etc. 

For  any  two  corresponding  points,  as  /  and  g,  these  arcs  are 
of  the  same  radius  and  are  struck  from  the  same  center ;  hence 


---dT' 


'Fig.  59  (Repeated) 

corresponding  parts  have  equal  elevations.  Therefore  the 
points  2^,  \^,  etc.,  are  real  points  of  intersection  for  these  arcs. 
Before  the  rotation  was  started,  the  lines  hf  and  })g  had  the  point 
h  in  common.  If  the  points  /  and  g  have  been  brought  together 
at  1,  these  lines  coincide  throughout. 

The  V  projection  of  the  points  1,  2,  etc.,  can  be  easily  found 
by  following  the  rotation  as  projected  on  a  V  plane  parallel 
to  the  plane  of  the  arc.  In  this  way  we  can  determine  the 
heights  of  each  of  the  required  points,  e,  h,  etc.  When  all  the 
polygons  are  thus  rotated  into  position,  the  cup-shaped  figure 


V,  §  52] 


PLANE   FIGURES  AND   SOLIDS 


61 


shown  by  the  soHd  lines  in  Fig.  59  is  obtained.  This  is  the 
lower  part  of  the  dodecahedron.  The  upper  part  is  found  by 
rotating  the  faces  downward,  making  a  cap-shaped  figure  as 
shown  by  solid  lines  on  Fig.  60.  This  figure  shows  the  com- 
plete projection  of  the  solid  found  by  placing  the  cap  on  the  cup. 


Fig.  60 


52.  Prisms.  A  right  prism  may  be  defined  as  a  solid  gen- 
erated by  the  motion  of  a  plane  polygon  which  moves  without 
rotating,  in  a  direction  perpendicular  to  its  own  plane.  If  this 
generating  polygon  is  a  regular  polygon,  the  prism  is  called 
regular,  and  the  line  through  the  center  of  the  polygon  perpen- 
dicular to  its  plane  is  called  the  axis  of  the  prism.  Then  we  may 
say  that  the  generating  polygon  always  remains  perpendicular 
to  the  axis  of  the  prism ;  that  any  plane  perpendicular  to  the 
axis  cuts  from  the  prism  a  section  equal  to  the  generating 
polygon,  and  that  no  other  plane  will  cut  out  of  the  prism  a 
section  exactly  equal  to  that  one. 

In  what  follows,  we  shall  be  interested  chiefly  in  regular 
prisms.  In  other  cases,  however,  any  line  perpendicular  to 
the  generating  polygon  may  be  substituted  for  the  axis  in  the 
preceding  statements. 

Since  a  section  made  by  a  plane  perpendicular  to  the  axis  is 
unique,  it  is  used  to  describe  the  prism.  Such  a  section  is  called 
a  right  section,  or  sometimes  simply  the  section,  of  the  prism. 


62 


DESCRIPTIVE   GEOMETRY 


[V,  §  53 


53.  To  Draw  the  Projections  of  a  Right  Prism.  If  it  is  desired 
to  draw  the  projections  of  a  right  prism,  the  right  section  as 
well  as  the  length  and  the  slope  of  the  axis  must  be  given.     The 


'.  b'      b'^ 


Fig.  61 


V,  §  53]  PLAXE   FIGURES  AXD   SOLIDS  63 

projections  of  the  axis  can  be  constructed  from  the  principles 
of  §  29,  and  then  the  projections  of  the  ends  can  be  found  from 
§  46.  Finally,  if  the  edges  are  drawn  to  connect  the  ends,  the 
projections  will  be  complete. 

For  example,  let  it  be  required  to  draw  the  projections  of  an 
hexagonal  prism,  f  in  diameter  and  H"  long,  with  its  axis 
inclined  at  angles  of  30°  and  45°  to  H  and  T",  respectively. 

In  Fig.  61,  aW'  and  a^'h^'  are  the  projections  of  such  a  line, 
determined  by  the  method  of  §  29,  as  shown  in  the  half-scale 
drawing  {a).  On  the  full-scale  drawing,  choose  an  auxiliary 
plane  of  projection  V  parallel  to  this  line,  as  indicated  by  the 
axis  line  H' A' .  Find  the  projection  of  the  axis  of  the  prism 
on  this  plane  (§  27),  as  shown  by  a^'^b^'. 

Now  the  bases  of  the  prism  will  be  planes  perpendicular  to 
the  new  plane  of  projection.  Hence  they  will  project  on  it 
in  lines  perpendicular  to  a^'h'' .  Draw  such  lines  of  indefinite 
length  {cd  and  cf) .  It  is  now  necessary  to  fix  the  definite  points 
on  this  line  which  represent  the  corners  of  the  hexagon.  In 
order  to  avoid  confusing  the  drawing,  imagine  that  the  generat- 
ing hexagon  has  been  slipped  out  along  the  axis  till  its  center  is 
at  x''\  Next  let  it  be  rotated  till  it  is  parallel  to  V\  Then  its 
projection  can  be  drawn  as  shown  at  1-6. 

By  projecting  these  points  back  to  the  lines  cd  and  ef  we  can 
locate  the  projections  of  the  required  corners  at  l^'-6^''  and 
'jvf_i2vf  'pj^p  fj  projections  of  these  points  lie  on  lines  drawn 
through  the  I  '  projections  and  perpendicular  to  H' A' .  Since 
the  prism  has  been  so  placed  that  the  points  6  and  2,  also  3 
and  5,  lie  on  lines  parallel  to  H,^  the  projection  2''6''  will  be 
equal  to  2-6  and  symmetrical  with  respect  to  the  projection 
of  the  axis,  a^h^.  Thus  the  plan  is  easily  completed.  The 
elevation  on  V  can  be  constructed  from  the  points  thus 
found  on  plan  together  with  the  heights  shown  on  the  V 
elevation. 

^  The  more  general  case  is  given  in  Art.  97. 


64 


DESCRIPTIVE   GEO:\IETRY 


[V,  §54 


54.  Intersection  of  Right  Prisms  —  Special  Case.^  When 
two  prisms  are  drawn  as  shown  in  Fig.  62  a,  it  is  not  at  once 
apparent  where  they  intersect.  The  problem  of  determining 
the  Hne  of  intersection  becomes  very  simple  when  it  is  re- 
membered that  the  line  of  intersection  will  be  that  line  which 
joins  the  points  where  the  edges  of  one  solid  pierce  the  faces 
of  the  other. 

Since  one  solid,  in  Fig.  62  a,  is  composed  entirely  of  vertical 
and  horizontal  planes,  the  obvious  method  is  to  find  the  points 
(«)  b*  (Jb) 


Fig.  62 


in  which  each  edge  of  the  inclined  prism  intersects  some  face  of 
the  upright  prism,  using  the  principle  of  §  31.  In  this  case 
we  find  the  trace  of  a  line  on  the  plane  bounding  the  prism 
instead  of  finding  its  trace  on  H  or  V. 

In  Fig.  62  a,  the  trace  of  the  edge  cd  is  found  by  noticing  that 
the  intersection  of  c^d^  and  ^^5^  is  the  point  k^.  This  point  is 
the  H  projection  of  the  trace  of  cd  on  the  plane  4-5-11-10. 
Drawing  a  line  through  k^  perpendicular  to  HA  till  it  intersects 

^  The  more  general  case  is  given  in  §  104. 


V,  §  54]  PLANE   FIGURES  AND   SOLIDS  65 

c^'d",  we  find  the  T^  projection  of  A*,  which  is  the  trace  of  cd  on 
the  prism.  By  the  same  method,  the  traces  of  the  other  lines 
may  be  found.  When  these  traces  are  properly  joined,  the 
intersection  as  shown  in  Fig.  62  b  is  determined. 

EXERCISE  SHEET  XVIII 

Take  P  at  the  left  border.     No  profile  projections  are  required. 

1.  Assume  a  card  in  the  shape  of  a  hollow  rectangle  1"  wide  and 
11"  long.  The  inner  rectangle  is  |"  wide  and  IJ"  long.  This  figure 
is  placed  with  one  corner  at  the  point  a  (8,  0,  12).  The  short  side  ah 
runs  forward  and  to  the  right.  The  Une  ab  lies  in  H  and  makes  an 
angle  of  30°  with  HA.  The  plane  of  the  figure  inclines  toward  V  and 
makes  45°  with  H.     Draw  its  H  and  V  projections. 

2.  Draw  the  H  and  V  projections  of  a  regular  tetrahedron  2"  on 
a  side,  with  its  base  parallel  to  H.  The  back  edge  of  the  base  is  the 
line  joining  the  points  i  (52,  4,  5)  and  j  (68,  4,  5). 

EXERCISE   SHEET   XIX 

Place  the  sheet  with  the  long  dimension  vertical,  and  with  the  wide 
margin  at  the  left.  Locate  HA  at  a  distance  3"  from  upper  border 
line.  Draw  the  H  and  V  projections  of  a  solid  bounded  by  6  squares 
and  8  equilateral  triangles.  Each  side  of  each  of  these  squares  and  of 
each  of  the  triangles  is  1"  long.  The  sohd  rests  on  a  triangle  as  a 
base,  the  center  being  placed  at  the  center  of  the  lower  part  of  the  sheet. 

EXERCISE    SHEET    XX 

Take  P  at  the  left.     No  profile  projections  are  required. 

1.  The  point  a  (20,  0,  12)  is  the  center  of  the  base  of  a  right  hexag- 
onal prism,  2|"  in  diameter  and  If"  high.  The  point  b  (17,  0,  14) 
is  the  center  of  the  base  of  a  regular  tetrahedron,  3"  on  a  side.  Find 
the  line  of  intersection  of  the  solids. 

2.  Assume  an  irregular  quadrilateral  prism  with  its  faces  vertical, 
and  an  irregular  triangular  prism  whose  edges  are  normal-case  lines. 
Find  the  fine  of  intersection  of  the  sohds.  The  exact  layout  is  left  to 
the  student. 

[Note.  Having  completed  the  foregoing  work,  the  student  can 
profitably  take  up  the  problems  of  §§  179-183  and  202-210.] 


CHAPTER  VI 
PROBLEMS   DEALING   WITH   POINTS,   LINES,   AND   PLANES 

55.  Introduction.  The  preceding  chapters  cover  the  essential 
principles  of  orthographic  projection.  By  the  use  of  these 
principles  only,  a  large  part  of  all  ordinary  drafting  operations 
may  be  performed.  The  great  majority  of  cases  deal  with 
objects  composed  of  plane  surfaces,  and  the  resulting  intersec- 
tions, which  are  straight  Hues  and  points. 

Cases  arise  constantly,  however,  in  which  it  is  necessary  to 
establish  relations  of  parallelism,  perpendicularity,  etc.,  between 
lines  and  planes,  and  to  record  the  results  in  projection.  The 
present  chapter  will  be  devoted  to  the  further  study  of  points 
and  lines  with  reference  to  their  relations  to  planes,  and  to  the 
representation  of  plane  surfaces  and  their  intersections. 

56.  Representation  of  Planes.  In  making  working  drawings, 
the  planes  dealt  with  are  usually  definite  in  extent  and  are  repre- 
sented by  the  projections  of  their  bounding  lines  or  points.  Thus, 
in  Fig.  7,  the  plane  A  is  defined  by  the  lines  ab,  be,  cd,  and  da. 

It  is  frequently  necessary,  however,  to  extend  these  planes 
indefinitely.  Hence  we  must  start  with  the  conception  of 
planes  of  indefinite  extent,  and  we  must  find  a  means  of  rep- 
resentation which  can  be  used  in  connection  with  the  methods 
of  point  and  line  projection  which  we  have  studied  before. 

From  solid  geometry  we  know  that  a  plane  is  determined : 
(1)  by  any  three  points,  and  (2)  by  any  two  intersecting  or 
parallel  lines.  Figure  63  a  shows  a  part  of  a  plane,  X,  and  three 
points,  a,  b,  and  c,  lying  in  X.  The  projections  of  these  points 
might  be  said  to  determine  the  plane,  X.  No  other  plane 
could  be  determined  by  these  projections,  but  any  other  three 

66 


VI,  §  56] 


POINTS,  LINES,  AND  PLANES 


67 


points  chosen  at  random,  in  X,  not  on  the  same  straight  line, 
would  determine  the  plane  equally  well.  Hence  the  projections 
of  three  points  do  not  constitute  an  entirely  satisfactory  de- 


scription of  the  plane,  since  the  choice  of  these  points  is  not 
unique.  In  the  same  way,  it  can  be  shown  that  the  projections 
of  intersecting  or  parallel  lines,  lying  in  X,  would  not  constitute 
a  unique  description. 
However,  if  the  plane 
is  considered  as  of  in- 
definite extent,  it  will 
intersect  the  planes  of 
projection  in  two  lines 
(Fig.  63  6,  XX\  XX'*), 
which  do  constitute  a 
unique  description  of 
the  plane.  These  lines 
are  called  the  traces  of 
the  plane.  (Compare 
this  with  the  definition 
of  the  traces  of  a  line,  §  31.)  Since  the  plane  is  indefinite  in 
extent,  it  will  extend  into  all  four  quadrants  (Fig.  64).  The 
traces  below  and  behind  HA  are  merely  continuations  of  those 
above  and  in  front  of  HA. 


Fig.  64 


68 


DESCRIPTIVE   GEOMETRY 


[VI,  §  56 


WTien  the  planes  of  projection  are  opened  or  flattened  out, 
these  traces  become  merely  two  lines  intersecting  on  HA,  as 

V 


Fig.  65 


Fig.  66 


shown  in  Fig.  65.  When  a  profile  plane  of  projection  is  used 
there  will  be  a  third  trace,  which  intersects  the  other  traces, 
one  on  each  of  the  profile  axes.     (See  Figs.  66  and  67.)     The 


Fig.  67 


profile  trace  is  little  used,  however.  Hereafter,  unless  otherwise 
specially  noted,  the  phrase  traces  of  a  plane  will  be  taken  to 
mean  the  traces  on  the  H  and  V  planes. 


VI,  §  59] 


POINTS,   LINES,  AND  PLANES 


69 


57.  Assuming  a  Plane.  It  will  be  evident  at  once  that  any 
two  lines  intersecting  on  HA  represent  the  traces  of  some 
plane ;  hence  to  assume  a  plane  it  is  sufficient  to  draw  any  two 
lines  intersecting  on  HA. 

It  may  be  well  to  note,  however,  as  a  precautionary  measure 
to  guide  one's  sense  of  visualization,  that  the  angle  X^  X  X^, 
in  Fig.  65,  is  not  equal  to  the 
corresponding  angle  in  Fig. 
64,  that  is,  to  the  one  which 
exists  in  space.  This  point 
is  brought  out  more  fully  in 
Fig.  68,  which  is  drawn  to 
show  the  opening  of  the 
planes  with  reference  to 
these  angles. 

58.  Normal    and    Special 
Cases.      For    our    purposes 

special-case  planes  may  be  defined  as  those  that  are  perpendicu- 
lar or  parallel  to  a  plane  of  projection.  It  will  be  a  helpful 
exercise  for  the  student  to  draw  the  traces  of  as  many  special-case 
planes  as  occur  to  him.  In  doing  this  start  with  the  limiting 
cases,  viz.,  (1)  when  the  plane  is  perpendicular  to  HA,  and  (2) 
when  the  plane  contains  HA.  Draw  also  the  traces  of  the 
normal  cases  which  are  in  nearly  the  same  position  as  (1)  and 
(2)  above ;  and  of  several  intermediate  cases.  Also  study  each 
special  case  with  reference  to  possible  variations  and  limits. 

59.  The  Use  of  Isometric  Projections.  In  the  study  of  plane 
problems,  the  power  of  visualization  is  heavily  taxed.  A  free 
use  of  isometric  sketches  similar  to  Figs.  63,  64,  and  67  will  be 
remarkably  helpful  in  this  respect.  In  the  end,  it  will  be  found 
that  the  making  of  such  sketches  has  created  the  ability  to 
visualize  without  such  help.  For  this  purpose  quickly  drawn 
freehand  sketches  are  quite  satisfactory  if  they  are  kept  in 
fairly  correct  proportion. 


70 


DESCRIPTIVE   GEO:\IETRY 


[VI,  §  60 


60.  The  Fundamental  Relations  of  Point,  Line,  and  Plane. 
A  point  can  be  represented  by  its  projections  only  (§7).  A  line 
may  be  represented  by  its  projections,  or  by  its  traces  (§§  14, 
15,  and  31).  A  plane  can  be  represented  satisfactorily  only  by 
its  traces  (§  56).  These  facts  should  be  clearly  recognized, 
since  many  of  the  important  principles  and  methods  of  the 
following  problems  depend  upon  them. 

Figure  69  shows  a  point  a,  on  the  plane  X,  and  Figure  70 
shows  the  same  point  in  projection.     For  normal-case  points. 


Fig.  69 


Fig.  70 


on  normal-case  planes,  the  projections  of  the  points  do  not  fall 
on  the  traces  of  the  planes.  Moreover,  there  is  no  simple  re- 
lation between  the  projections  of  any  point  and  the  traces  of 
the  plane  which  contains  it,  beyond  the  fact  that  a^  (Fig.  70) 
must  lie  within  the  angle  X""  XA,  and  a!"  must  he  within  the  angle 
AXX^.  Let  the  student  work  out  this  idea  in  its  application 
to  points  in  the  second,  third,  and  fourth  quadrants. 

In  certain  special  cases,  one  or  both  of  the  projections  of  a 
point  may  lie  on  the  traces  of  the  plane.  Let  the  student  work 
out  a  few  such  cases. 


VI,  §  60]  POINTS,   LINES,  AND  PLANES 


71 


Figures  71  and  72  show  the  projections  of  a  normal-case  line, 
aby  lying  in  a  normal-case  plane,  Y.  As  long  as  the  line  is 
limited,  and  is  described  merely  by  means  of  the  normal-case 


Fig.  71 


points  a  and  h,  no  particular  relations  appear  to  exist  between  the 
traces  of  Y  and  the  projections  of  ab.  However,  the  traces 
of  the  line,  m  and  n,  being  special-case  points,  do  have  a  definite 


relation  to  the  traces  of  the  plane.  It  will  be  seen  that  the  H 
trace  of  the  line  must  fall  in  the  H  trace  of  the  plane  ;  and  the 
V  trace  of  the  line  must  fall  on  the  V  trace  of  the  plane.     Why  ? 


72 


DESCRIPTIVE   GEOMETRY 


[VI,  §  61 


61.   Special  Cases.     Special-case  lines  or  special-case  planes 
may  show  definite  relations  between  the  projections  of  the  line 


Fig.  73 
and  the  traces  of  the  plane.     One  such  case  is  of  sufficient 


Fig.  74 


importance  to  deserve  particular  notice.     Figure  73  shows  a  line 
lying  in  the  plane  Z,  and  parallel  to  //.     It  can  be  proved  readily 


VI,  §  63] 


POINTS,   LINES,  AND  PLANES 


73 


that  the  H  projection  of  such  a  line  is  parallel  to  the  H  trace 
of  the  plane,  and  that  the  V  projection  of  the  line  is  parallel  to 
HA,  as  shown  in  Fig.  74. 

Let  the  student  prepare  a  proof  of  the  above  statement,  and 
work  out  the  other  special  cases  in  which  there  are  definite 
relations  between  the  projections  of  the  line  and  the  traces  of 
the  plane. 

62.  Use  of  the  Profile  Plane.  In  Fig.  75,  let  the  lines  ah 
and  cd  be  given,  and  let  the  traces  of  the  plane  determined  by 


VI 

x" 

X' 

a" 

b^ 

\ 

i)a''b'* 

nf 

\' 

c" 

d^ 

\cMf 

1  1 

A 

nr\ 

a^  i 

b^ 

/            ' 

/ 

Ic*' 

d^ 

• 

X. 

x^ 

p 

Fig.  75 


them  be  required.  The  given  lines  have  no  H  and  V  traces ; 
hence  it  is  necessary  to  use  a  profile  plane. 

Find  the  profile  traces  of  the  lines  {m.  and  ^0-  The  profile 
trace  of  the  required  plane  will  pass  through  these  points,  and 
its  intersections  with  the  profile  axes  ( Xi  and  X^)  will  determine 
the  required  H  and  V  traces,  X^X^  and  X'X". 

63.  Choice  of  Auxiliary  Lines  and  Planes.  In  the  problems 
that  follow,  it  will  be  necessary  to  make  frequent  use  of  auxiliary 


74 


DESCRIPTIVE   GEOMETRY 


[VI,  §  63 


lines  and  planes,  i.e.  lines  or  planes  arbitrarily  introduced  into 
the  problem  in  order  to  arrive  at  a  solution.^ 

For  the  purposes  of  §§  65  and  67,  where  an  infinite  number  of 
solutions  is  possible,  any  one  of  an  infinite  number  of  auxiliary 
lines  may  be  chosen.  When  the  conditions  of  the  problem  are 
prescribed  more  and  more  closely,  however,  it  becomes  more 
and  more  necessary  to  use  good  judgment  in  choosing  the 
auxiliary  lines  and  planes  so  that  the  more  rigid  conditions  of 
these  problems  may  be  fulfilled.  When  only  one  solution  of 
a  problem  is  possible,  the  choice  of  an  auxiliary  becomes  quite 
restricted.  Moreover,  the  judicious  choice  of  auxiliary  lines 
and  planes  often  may  simplify  materially  the  necessary  drawing. 
Such  a  case  is  treated  in  §  78. 

It  follows  that  the  student  should  make  a  careful  study  of  the 
way  in  which  his  choice  affects  any  given  solution,  and  thus 
strengthen  his  ability  to  select  instinctively  the  best  possible 
auxiliary  lines  and  planes  for  any  given  problem. 

64.  To  Assume  a  Line  Which  Lies  in  a  Given  Plane.  In  Fig. 
76,  let  the  plane  Z  be  given  and  let  it  be  required  to  assume  a 
line  which  lies  in  Z. 

i 


Fig.  76 

^  This  idea  of  arbitrarily  introducing  a  foreign  element  into  a  problem 
in  order  to  arrive  at  a  solution  is  frequently  used  in  mechanics,  biology, 
chemistry,  medicine,  and  other  sciences,  as  well  as  in  all  branches  of  pure 
mathematics. 


VI,  §  66] 


POINTS,   LINES,   AND  PLANES 


75 


Since  the  traces  of  any  such  line  are  known  to  He  in  the  traces 
of  the  plane,  any  two  points  chosen  at  random,  one  on  ZZ^^ 
the  other  on  ZZ"^,  will  be  the  traces  of  a  line  in  Z.  Let  the  point 
m  be  chosen  on  ZZ^'.  This  point  lies  in  V  \  therefore  its  // 
projection  lies  in  E.A,  at  m^.  In  the  same  way  n^  and  iV^  are 
the  projections  of  some  point  in  the  H  trace  of  Z,  By  con- 
necting tny-ny  and  n^m^^  we  may  determine  the  projections  of  the 
line  mriy  which  lies  in  Z.  • 

65.  To  Assume  a  Point  Which  Lies  in  a  Given  Plane.  This 
problem  is  complicated  by  the  fact  mentioned  in  §  60,  that  a 
point  has  projections  only,  whereas  a  plane  has  traces  only. 
For  this  reason  it  is  necessary  to  employ  an  auxiliary  line. 
The  line,  ha\ing  both  projections  and  traces,  can  be  used  to 
establish  relations  between  the  point  and  the  plane. 

It  is  sufficient,  then,  to  assume  any  line  in  the  plane  (§  64). 
Any  point  on  such  a  line  will  lie  in  the  given  plane.  For  this 
purpose,  such  a  line  as  is  shown  in  Fig.  73  is  frequently  found 
most  convenient. 

66.  To  Pass  a  Plane  through  a  Given  Line.  In  Fig.  77,  let 
ah  be  the  given  line.     Determine  its  traces,  c  and  d.     Through 


FiQ.  77 


these  points  draw  any  two  lines  that  intersect  on  HA  (XX^ 
and  XX^).     These    lines   will   be   the   traces   of  some   plane 


76 


DESCRIPTIVE   GEOMETRY 


[VI,  §  60 


passing  through  ab.     An  infinite  number  of  such  planes  can 
be  drawn. 

67.  To  Pass  a  Plane  through  a  Given  Point.  Pass  an  auxil- 
iary line  through  the  given  point,  then  pass  a  plane  through  the 
line  (§  66).  This  plane  will  contain  the  given  point.  Here 
again  the  line  parallel  to  a  plane  of  projection  is  usually  most 
convenient  (Fig.  73). 

68.  To  Find  the  Traces  of  the  Plane  Determined  by  Two 
Given  Intersecting  Lines.     Let  the  lines  ab  and  cd,  in  Fig.  78, 


Traces  of  a  Plane  Determixed  by  Two  Intersecting  Lines 

Fig.  78 


be  given  and  let  it- be  required  to  find  the  traces  of  the  plane 
determined  by  them.     It  should  be  noticed  that  the  lines  must 


VI,  §  68]  POINTS,   LINES,  AND  PLANES 


77 


intersect  in  order  to  determine  a  plane  (§  33).     The  given  lines 
intersect  at  e. 

The  traces  of  the  given  lines  will  lie  in  the  required  traces  of 
the  planes  (§  60) ;  therefore  if  the  traces  of  ab  {n  and  q)  and  of 
cd  {m  and  y)  are  found  (§  31),  these  traces  will  determine  the 
traces  of  the  required  plane  X.  If  the  work  has  been  done 
carefully,  the  lines  thus  determined  will  meet  on  HA.  This 
furnishes  a  test  of  accuracy  of  draftsmanship. 


Isometric  Projection  Corresponding  to  Fig.  78 
Fig.  79 


Figure  79  shows  the  space  figure  that  corresponds  to  Fig.  78, 
in  isometric  projection. 

Sometimes  it  may  be  inconvenient  or  impossible  to  find  all 
four  traces  of  the  given  lines.  In  this  event,  determine  two 
traces  on  either  H  or  V ;   draw  the  trace  of  the  plane  through 


78  DESCRIPTIVE   GEOMETRY  [VI,  §  68 

these  points  and  continue  it  to  HA.  The  remaining  trace  of 
the  line  together  with  the  point  just  found  on  HA  will  determine 
the  other  trace  of  the  plane. 

69.  Plane  Determined  by  Three  Points.  Let  any  three  points 
in  space  be  given,  and  let  it  be  required  to  find  the  traces  of  the 
plane  determined  by  them.  It  is  sufficient  to  draw  two  inter- 
secting lines  through  the  three  points  and  proceed  as  above. 
However,  since  the  three  points  determine  three  lines,  any  two 
of  which  intersect,  some  room  is  left  for  the  exercise  of  good 
judgment  in  selecting  the  pair  that  will  give  the  easiest  solution. 

EXERCISE   SHEET   XXI 

Take  P  at  the  left,  20  from  the  border  line,  and  turned  away  from 
the  objects,  into  V. 

1.  Draw  the  H,   V,  and  P  traces  of  the  plane  R  ^  (22,  72"  45°  I, 

2.  Given  the  points  a  (IV,  33,  20,  2),  b  (IV,  54,  20,  2),  c  (IV,  33, 
4,  7),  and  d  (IV,  54,  4,  7).  Find  the  H,  V,  and  P  traces  of  plane  S 
determined  by  the  Hnes  ab  and  cd. 

EXERCISE   SHEET   XXII  ^ 

[Note.  Unless  otherwise  indicated,  P  will  hereafter  be  taken  at 
the  right  border  line  and  no  profile  projections  will  be  required.] 

1.  Given  the  plane  R  (75,  i^'' 45°  r,  R""  90°),  draw  the  projections 
of  two  Hnes  lying  in  R ;  one  parallel  to  and  1 1  from  H,  the  other  parallel 
to  and  6  from  V.     Letter  these  lines  ab  and  cd. 

2.  Given  the  plane  S  (31,  Sf"  30°  I,  S"  45°  I),  locate  a  point  e  on  S 
and  at  a  distance  7  from  H  and  3  from  V. 

3.  Given  the  points/  (18,  9,  4)  and  g  (9,  3,  12).  Through /gr  pass 
three  planes,  T,  U,  and  W,  the  plane  T  being  perpendicular  to  H. 

1  For  explanation  of  the  notation  used  in  describing  planes,  see  Descrip- 
tion of  Quantities,  page  xiii. 

2  Sheets  XXII  to  XXXVII  consist  of  three  problems  each,  so  spaced  that 
#1  from  any  sheet  will  combine  readily  with  #2  and  #3  from  any  other  sheet 
or  sheets. 


VI,  §  71]  POINTS,   LINES,   AND  PLANES  79 

EXERCISE   SHEET   XXni 

1.  Given  the  points  a  (III,  72,  4,  11)  and  6  (57,  6,  5).  Through  ab 
pass  the  plane  X  intersecting  HA  at  70. 

2.  Given  the  points  c  (48,  6,  10)  and  d  (37,  6,  2).  Through  cd  pass 
the  plane  Y  intersecting  HA  at  29. 

3.  Given  the  point  e  (15,  8,  6).  Through  e  pass  the  plane  Z  meet- 
ing HA  at  3. 

EXERCISE   SHEET   XXIV 

1.  Through  the  point  a  (65,  5,  4)  pass  two  lines  as  follows.  The 
H  projection  of  one  line  makes  an  angle  of  30°  to  the  left  and  the  V 
projection  an  angle  of  60°  to  the  right  with  HA.  The  H  projection 
of  the  other  makes  an  angle  of  60°  to  the  right  and  the  V  projection 
an  angle  of  45°  to  the  left  with  HA.  Draw  the  traces  of  the  plane  X 
determined  by  these  hnes. 

2.  Given  the  line  joining  the  points  h  (46,  11,  2)  and  c  (34,  3,  14), 
and  the  point  d  (II,  41,  2,  5).  Draw  the  traces  of  the  plane  Y  deter- 
mined by  the  point  d  and  the  line  ab. 

3.  Given  the  points  e  (23,  2,  12),  /  (20,  11,  4),  and  g  (14,  2,  4). 
Draw  the  traces  of  the  plane  of  the  triangle  efg. 

70.  Parallel  Planes.  If  two  planes  are  parallel,  their  cor- 
responding traces  are  parallel.  Let  the  student  prove  this  by 
using  the  theorem  of  solid  geometry  concerning  the  lines  of 
intersection  of  two  parallel  planes  with  any  third  plane. 

71.  Line  Parallel  to  a  Plane.  In  general,  when  a  line  is  parallel 
to  a  plane,  the  projections  of  the  line  are  not  parallel  to  the 
traces  of  the  plane. 

Exceptions  to  this  statement  may  be  noted  in  the  cases  of 
planes  whose  traces  are  both  perpendicular  to  HA.  Another 
exception  is  the  case  in  which  both  the  line  and  the  plane  are 
parallel  to  HA. 

If  a  line  is  parallel  to  a  plane  and  also  to  H  or  V,  one  pro- 
jection of  the  line  w^ll  be  parallel  to  one  trace  of  the  plane.  This 
case  is  similar  to  that  shown  in  Fig.  73. 


80 


DESCRIPTIVE   GEOMETRY 


[VI,  §  72 


72.  To  Pass  a  Plane  through  a  Line  and  Parallel  to  Another 
Line.  From  solid  geometry  we  know  that  a  line  and  a  plane 
are  parallel  if  the  plane  contains  any  line  which  is  parallel  to 
the  given  line. 

In  Fig.  80,  let  the  lines  ah  and  cd  be  given.  Let  it  be  required 
to  pass  a  plane  through  ah,  parallel  to  cd.  Choose  any  point 
e,  on  ab,  and  through  it  draw  the  auxiliary  line  fg,  parallel  to 
cd.  The  two  lines  ah  and  fg  determine  the  plane  Z  (§  68). 
This  plane  is  parallel  to  cd,  since  it  contains  a  line  parallel  to  cd. 


Fig.  80 


73.  To  Pass  a  Plane  through  a  Point  and  Parallel  to  a  Given 
Line.  The  analysis  and  construction  are  so  nearly  identical  with 
that  of  §  72  that  no  further  explanation  is  needed.  The  student 
should  recognize,  however,  that  the  problem  is  indeterminate, 
as  are  those  of  §§  66  and  67. 

74.  To  Pass  a  Plane  through  a  Given  Point  and  Parallel  to  a 
Given  Plane.  The  traces  of  the  required  plane  will  be  parallel 
to  the  traces  of  the  given  plane  (§  70).  An  auxiliary  line,  pass- 
ing through  the  given  point,  is  needed  to  determine  the  required 
plane  (§  67).     In  this  case  the  problem  is  simplified  by  choosing 


VI,  §  741  POINTS,   LINES,   AND   PLANES 


81 


as  the  auxiliary  a  line  parallel  to   H  and  to  the  given  plane. 
Refer  to  §§  63,  61  and  the  last  paragraph  of  §  71. 

In  Fig.  81,  let  the  plane  Z  and  the  point  a  be  given,  and  let  it 
be  required  to  pass  a  plane  through  a,  parallel  to  Z.  Draw  the 
H  projection  of  the  auxiliary  line  through  a^  and  parallel  to 
ZZ^,  as  nhn^.  The  V  projection  of  such  a  line  (since  the  Hne  is 
parallel  to  H)  will  be  parallel  to  HA,  and  will  pass  through 


FiQ.  81 

a^,  as  shown  by  n'-'m''.  The  V  trace  of  ??m  (??')  will  lie  in  the 
V  trace  of  the  required  plane.  Hence  a  line  drawn  through 
71^  parallel  to  ZZ%  will  be  one  of  the  required  traces,  Y  Y". 
Through  7  draw  YY^  parallel  to  ZZ\ 

EXERCISE   SHEET   XXV 

1.  Given  the  plane  M  (75,  M^  60°  r,  M"  45°  r)  and  the  point  a 
(61,  6,  8).     Through  a  draw  a  line  parallel  to  M. 

2.  Given  the  plane  .V  (45,  A'^  75°  I,  N^  45°  r)  and  the  point  b  (39, 
10,  12).     Through  h  draw  a  line  parallel  to  N. 

3.  Given  the  points  c  (III,  25,  11,  19),  d  (III,  16,  15,  11),  e  (20,  2, 
14),  and  /  (8,  9,  4).     Through  the  line  ef  pass  a  plane  0  parallel  to  cd. 


EXERCISE   SHEET   XXVI 

1.   Given  the  points  g  (III,  77,  6,  17),  h  (III,  70,  6,  8),  j  (68,  1,  11), 
and  k  (60,  12,  3).     Through  the  line  jk  pass  a  plane  Q  parallel  to  gh. 


82 


DESCRIPTIVE   GEOMETRY 


[VI,  §  74 


2.  Given  the  points  I  (51,  3,  12),  m  (42,  8,  5),  and  n  (36,  7,  7). 
Through  n  pass  the  plane  R  parallel  to  the  Une  Im. 

3.  Given  the  points  q  (IV,  16,  7,  14),  r  (IV,  24,  18,  9),  and  o  (II, 
12,  9,  3).     Through  o  pass  the  plane  S  parallel  to  the  line  qr. 

EXERCISE    SHEET   XXVII 

1.  Given  the  plane  X  (80,  X^"  60°  r,  Z"  45°  r)  and  the  point  a  (63, 
7,  6).  Through  a  pass  the  plane  Y  parallel  to  X.  (Use  a  normal-case 
auxiliary  line.) 

2.  Given  the  plane  Z  (45,  Z^  30°  r,  Z^  60°  0  and  the  point  h  (III, 
38,  8,  7).  Through  b  pass  the  plane  W  parallel  to  Z.  (Use  a  special- 
case  line.) 

3.  Given  the  plane  U  (3,  m  30°  Z,  C/'^  45°  I)  and  the  point  c  (IV, 
16,  5,  7).     Through  c  pass  the  plane  T  parallel  to  U. 

75.  The  Line  of  Intersection  of  Two  Planes.  The  line  of 
intersection  of  two  planes  contains  all  points  common  to  the 
two  planes.     If  the  traces  of  two  planes  intersect,  the  point  of 


Fia.  82 


intersection  is  a  point  common  to  the  two  planes ;  hence  it  will 
be  in  the  line  of  intersection  of  the  planes.  Two  such  points 
are  necessary  to  fully  determine  the  line  of  intersection  of  any 
two  planes. 


VI,  §  76] 


POINTS,   LINES,   AND  PLANES 


83 


In  Fig.  82  are  shown  two  planes,  A^  and  Y.  All  points  on 
X  X"  lie  in  X.  All  points  on  Y  7^  lie  in  Y.  Therefore  the 
point  a,  which  lies  in  both  the  lines,  is  in  both  the  planes,  and 
hence  it  is  on  their  line  of  intersection.  The  same  is  true  for 
the  point  b. 

Figure  83  shows  the  same  operation  in  projection.  The  V 
traces  of   X  and    Y  are  extended  to  their  intersection  at  o^. 


H 


This  point  lies  in  V,  and  hence  a^  lies  on  HA.  In  the  same 
way  b^  and  b^  are  found  by  using  the  H  traces  of  the  given  planes. 
Hence  the  lines  a^b^  and  a"6''  are  the  projections  of  the  line  of 
intersection  of  the  planes  X  and   Y. 

[Note.  The  relation  between  such  a  line  as  ab,  above,  and  its 
projections  seems  to  be  difficult  for  the  average  student  to  visualize. 
Quite  a  common  mistake  consists  in  connecting  a*  and  b^  to  show  the 
line  of  intersection.  A  little  study  will  show  that  such  a  line  is  mean- 
ingless on  a  projection  drawing.  It  is  true  that  the  line  in  space  does 
connect  these  points;  but  in  the  projection  the  V  projection  of  one 
point  and  the  H  projection  of  another  cannot  determine  a  line.] 

76.  Special  Cases.  \Mien  the  traces  of  the  given  planes  do 
not  meet  within  the  limits  of  the  drawing,  the  method  of  §  75  for 


84 


DESCRIPTR^   GEOMETRY 


[VI,  §  76 


finding  the  line  of  intersection  cannot  be  applied  directly. 
However,  a  solution  usually  can  be  obtained  within  the  limits  of 
the  drawing,  by  using  a  well  chosen  auxiliary  plane. 

Case  L  In  Figs.  84  and  85,  the  plane  Z  is  taken  as  the 
auxiliary  plane  and  is  erected  parallel  to  V.  This  plane  con- 
stitutes, in  effect,  a  new  V  plane  on  which  the  traces  of  X  and 


FiQ.  84 

y  intersect  at  the  point  d.  These  traces,  when  projected  on 
the  original  V  plane,  will  intersect  within  the  limits  of  the 
drawing,  making  the  following  solution  possible. 

Referring  to  Fig.  85,  the  plane  Z  is  passed  parallel  to  F  and 
cutting  X  and  7.  The  points  h  and  c  are  seen  to  lie  on  the  Hues 
of  intersection  between  the  given  planes  and  Z.  The  V  pro- 
jections of  U  and  cd  will  be  parallel  to  Y  7^  and  XX-  (§  61). 
Thus  </^  is  established.     Now  d^  will  lie  on  ZZ^  below  d\     The 


VI,  §  761  POINTS,   LINES,   AND  PLANES 


85 


Fig.  85 


points  a  and  d  determine 
the  required  line  of  inter- 
section between  X  and  Y. 

Case  II.  If  neither  pair 
of  the  given  traces  intersect 
on  the  paper,  two  auxihary 
planes,  one  parallel  to  V, 
the  other  parallel  to  //,  may 
be  used  after  the  manner  of 
Case  I. 

Case  III.  It  is  some- 
times more  convenient  to 
use  an  auxiliary  plane  paral- 
lel to  one  of  the  given  planes, 
as  shown  in  Fig.  86. 

Here  Z  is  passed  parallel 
to   X,  and  intersects   7  in  a  line,  hdy  which  is  parallel  to  the 
required  line  of  intersection  ac,  and  which  will  have  its  projec- 
tions lying  wholly  within  the  limits  of  the  drawing.     Let  the 

student  make  the  con- 
struction in  projection. 
This  solution  is  particu- 
larly valuable  in  cer- 
tain cases  where  one 
plane  is  parallel  to  HA 
and  a  profile  projec- 
tion is  not  easily  con- 
structed. 

Case  IV.  If  both  of 
the  given  planes  are 
parallel  to  HA,  as 
shown  in  Fig.  87,  and 
if  a  profile  projection 
Fig.  86  '  is      not      wanted,     an 


86 


DESCRIPTIVE   GEOMETRY 


[VI,  §  76 


auxiliary  plane  cutting  both  of  the  given  planes  may  be  used. 
In  the  figure,  the  plane  Z  cuts  X  in  the  line  ah  and  it  cuts  1'  in 
the  line  cd  (§  75).  These  lines  of  intersection  will,  themselves, 
intersect  (since  they  are  in  the  same  plane),  giving  one  point, 
e,  on  the  required  line  of  intersection.  The  required  line  will 
pass  through  c  and  be  parallel  to  HA. 

n.   Use  of  the  Auxiliary  Plane.     The  solution  for  Case  1  of 
76  is  more  than  a  mere  expedient  for  solving  a  single  problem. 

-r    Y 


x^ 

oV 

^ 

/ 

X 

r^ 

fp"* 

T  • 

^•i 

Xb'     d^ 

A 

X     i        \ 

ejU--^ 

g*' 

1      ^V:  ^--\ 

YK 

c'> 

\ 

> — 

Y 
X 

Fig.  87 

Note  that  by  choosing  the  auxiliary  plane  parallel  to  V,  we  have, 
in  effect,  merely  shifted  V  forward,  temporarily,  in  order  to  fix 
points  otherwise  beyond  our  reach. 

The  principle  involved  can  best  be  understood  by  making 
a  sketch  like  Fig.  85.  Fold  this  sketch  as  shown  in  Fig.  88. 
In  this  manner  HA  is  brought  down  to  ZZ^,  eliminating  from  the 
drawing  all  the  space  between.  Disregarding  the  lines  YY^ 
and  XX^,  it  will  be  seen  that  the  V  plane  of  projection  has  been 
brought  forward  to  the  position  occupied  by  Z  in  Fig.  84,  and 
that,  by  using  the  axis  line  thus  established,  the  problem  can 
be  solved  by  the  method  of  §  75. 


VI,  §  77]  POINTS,   LINES,   AND  PLANES  87 

Cases  2  and  3  are  merely  slight  modifications  of  the  above 
principle. 

This  device  of  temporarily  shifting  the  position  of  one  or 
both  of  the  planes  of  projection  in  order  to  confine  a  solution 


Fig.  88 


IS 


to  a  limited  space  is  of  great  convenience  in  many  cases.     It  i 
sometimes  spoken  of  as  ''  shifting  the  axis  fine  "  or  *'  choosing 
a  new  plane  of  projection." 

EXERCISE   SHEET   XXVIU 

Find  the  lines  of  intersection  of  each  of  the  following  pairs  of  planes. 

1.  G  (76,  &  60°  r,  G''  45°  r)  and  H  (57,  m  30°  l,  H-  60°  /). 

2.  J  (50,  J"  45°  r,  J^'  90°^  and  K  (30,  /v"  60°  /,  K''  30°  /). 

3.  L  (18,  L^  30°  r,  L-  75°  /)  and  M  (8,  M^  60°  r,  M-  45°  r). 

EXERCISE    SHEET    XXIX 

Find  the  lines  of  intersection  of  each  of  the  following  pairs  of  planes. 

1.  N  (74,  A^''  45°  r,  N^  75°  I)  and  0,  parallel  to  HA  ;   O^  is  16  above 
and  0"  is  9  above  HA . 

2.  P  (50,  P^  821°  r,  P^'  75°  r)  and  Q  (28,  (?^  75°  I,  Q"  75°  Z). 

3.  R,  parallel  to  FA,  R^  is  5  below  and  R'  is  27  above  HA  ;    and 
S,  parallel  to  HA,  S^"  is  17  below  and  S''  is  8  above  HA. 

EXERCISE   SHEET   XXX 

Find  the  lines  of  intersection  of  each  of  the  following  pairs  of  planes. 

1.  T  (79,  Th  45°  r,  T"  75°  r)  and  U  (61,  T''  30°  /,  V  75°  r). 

2.  X   (49,   X''60°r,  XM5°  r)   and  TT,   parallel  to  //A.     JV^  is  6 
below  and  TT^^'  11  above  HA. 

3.  Y  (23,  7''  60°  r,  F''  90°)  and  Z  (3,  Z^  45°  ?,  Z^'  75°  0- 


88 


DESCRIPTIVE  GEOMETRY 


[VI,  §  78 


78.  To  Determine  the  Trace  of  a  Given  Line  on  a  Given 
Plane.  In  this  case,  of  course,  the  line  does  not  lie  in  the  plane. 
Hence  the  traces  of  the  line  on  H  and  V  bear  no  definite  relations 

to  the  traces  of  the  plane.  Thus, 
the  old  difficulty  of  establishing 
relations  between  a  line  (shown 
only  by  its  projections)  and  a 
plane  (shown  only  by  its  traces) 
is  encountered  again  (§  60).  In 
Fig.  89,  it  can  be  seen  merely  by 
visualization  that  the  plane  Z  and 
the  line  ah  are  not  parallel.  Hence 
they  must  Intersect.  But  the  point 
of  intersection  (the  trace  of  ah  on  Z) 
cannot  be  determined  so  easily. 
Here  again  the  solution  is  obtained  by  the  use  of  an  auxiliary 
plane;,  which  this  time  should  be  passed  through  the  line.  From 
Fig.  90,  which  is  a  Cavalier  projection  of  the  case  shown  in 


Fig.  90 


Fig.  89,  it  will  be  seen  that  the  line  of  intersection  of  the  auxiliary 
plane  Y,  with  the  given  plane  Z  (cd),  contains  the  required 
trace,  ?',  of  ah  on  Z. 


VI,  §  78]  POIXTS,   LINES,   AND  PLANES 


89 


Fia.  91 


In  Fig.  91,  the  operation  is  carried  out  in  projection.     The 

plane  Z  and  the  line  ah  are  given.     The  auxiliary  plane    Y  is 

passed  through  the  line 

ah  (§  66).     The  hne  of 

intersection  of  Z  and  Y , 

cd,  is  next  determined 

(§  75) .    This  Hne  is  seen 

to    intersect   the   given 

line  ah  at  i,  which  is  the 

required    trace    of    the 

line  ah  on  the  plane  Z. 
It  will  be  seen  that 

i^  and  i^  are  determined 

independently,  so  that 

if  the  construction  has 

been  accurately  drawn, 

they  should  lie  on  the  same  line  perpendicular  to  HA. 
By  a  judicious  choice  of  the  auxiliary  plane,  the  construction 

often  can  be  simplified  materially.     In  Fig.  92,  using  the  same 

line  and  plane  as  be- 
fore, the  auxiliary  plane 
has  been  passed  through 
ah,  and  perpendicular 
to  H.  The  H  trace, 
IT'',  will  pass  through 
a^h^,  while  the  V  trace, 
YY'\  will  be  perpen- 
dicular to  HA.  Thus 
the  determination  of 
the  auxiliary  plane  is 
greatly  simplified.  Also 
one  projection  of  the 
line  of  intersection  between  Z  and  Y,  c^d^,  will  coincide  with 
a^h^.     The  V  projection  gives  the  required  point  i. 


Fig.  92 


90  DESCRIPTIVE  GEOMETRY  [VI,  §  78 

By  comparing  Figs.  91  and  92,  the  simplification  brought 
about  by  choosing  an  auxiHary  plane  which  is  perpendicular 
to  a  plane  of  projection  will  be  evident.  Sometimes  it  is  better 
to  choose  the  auxiliary  plane  perpendicular  to  one,  and  some- 
times to  the  other  of  the  projection  planes.  The  proper  choice 
will  depend  upon  the  relative  positions  of  the  line  and  the  plane. 

Applications  of  this  problem  will  be  found  in  §  188. 

EXERCISE   SHEET   XXXI 

1.  Find  the  trace  of  the  hne  joining  the  points  a  (72,  2,  12)  and 
6  (62,  9,  5)  on  the  plane  Z  (58,  Z^  60°  /,  Z''  45°  T).  Use  a  normal-case 
auxihary  plane. 

2.  Find  the  trace  of  the  line  joining  the  points  c  (46,  12,  7)  and 
d  (III,  31,  9,  3)  on  the  plane  Y  (49,  Y^  45°  r,  7"  60°  r).  Use  a  special- 
case  auxiliary  plane. 

3.  Find  the  trace  of  the  line  joining  the  points  e  (20,  10,  7)  and 
/  (8,  2,  7)  on  the  plane  X  (18,  X^  60°  I,  X-  30°  r). 

EXERCISE   SHEET   XXXH 

Take  P  at  the  right  border  hne,  turned  toward  the  objects,  into  V. 
Use  P  only  when  necessary.    * 

1.  Find  the  intersection  of  the  Hne  joining  the  points  g  (72,  5,  6) 
and  h  (60,  11,  10)  with  the  plane  W  (75,  W^  45°  r,  TF"  90°). 

2.  Find  the  intersection  of  the  Hne  joining  the  points  i  (46,  13,  10) 
and  j  (III,  32,  9,  2)  with  the  plane  U,  which  is  parallel  to  HA,  U^ 
being  6  below  and  U''  10  above  HA. 

3.  Find  the  intersection  of  the  line  joining  the  points  k  (20,  18,  9) 
and  I  (20,  2,  9)  with  the  plane  T,  which  passes  from  II  to  IV,  through 
HA.  and  makes  an  angle  of  60°  with  H. 

79.  Line  Perpendicular  to  a  Plane.  If  a  line  is  perpendicular 
to  a  plane,  the  projections  of  the  line  are  perpendicular  to  the 
traces  of  the  plane.  This  will  be  seen  by  reference  to  Fig.  93. 
The  line  ah  is  perpendicular  to  the  plane  7.  The  H  projecting 
plane  Z  is  shown  passing  vertically  downward  through  it.  This 
plane  Z  is  perpendicular  to  //  and  also  to  7,  since  it  contains 
ab,  which  is  perpendicular  to   7. 

Since  both  //  and   7  are  perpendicular  to  Z,  the  line  of  inter- 


VI,  §  81]  POINTS,   LINES,   AND  PLANES 


91 


section  between  H  and  Y,  Y  Y^,  is  perpendicular  to  Z,  and 
hence  to  every  line  in  Z  passing  through  the  point  c.  There- 
fore  Y  y*  is  perpendicular  to  a^b'^. 


Fia.  93 

By  the  same  reasoning  it  can  be  proved  that  YY""  is  perpen- 
dicular to  a'"6''. 

80.  Perpendicular  Planes.  Let  the  student  visualize  for 
himself  the  truth  of  the  following  statements  : 

(1)  "When  two  planes  are  perpendicular,  their  traces  are  not, 
in  general,  perpendicular. 

(2)  The  limiting  conditions  occur  (a)  when  the  line  of  inter- 
section of  the  planes  is  parallel  to  a  plane  of  projection,  —  in 
this  case  the  traces  on  that  plane  of  projection  are  parallel ; 
and  (6)  when  the  line  of  intersection  of  the  two  planes  is  per- 
pendicular to  a  plane  of  projection,  —  in  this  case  the  traces  on 
that  plane  of  projection  are  perpendicular. 

(3)  An  exception  to  the  preceding  statement  occurs  when  one 
of  the  planes  is  perpendicular  to  a  plane  of  projection,  as  in 
Fig.  93. 

81.  To  Pass  a  Plane  Perpendicular  to  a  Given  Line.  It  is 
merely  necessary  to  draw  the  traces  of  the  required  plane 
perpendicular  to  the  projections  of  the  given  line. 


u 


92  DESCRIPTIVE   GEOMETRY  [VI,  §  82 

82.  To  Pass  a  Plane  through  a  Given  Point  and  Perpendic- 
ular to  a  Given  Line.  Pass  any  plane  perpendicular  to  the 
given  line  (§  81).  Then  pass  a  plane  through  the  given  point, 
parallel  to  the  plane  just  drawn  (§  74). 

After  following  the  above  analysis  of  the  problem,  and  solving 
an  example  or  two,  the  student  will  doubtless  discover  a  way 
to  simplify  the  construction. 

83.  To  Pass  a  Plane  Parallel  to  a  Given  Plane  and  at  a  Given 
Distance  from  It.^  The  distance  between  two  parallel  planes  is 
measured  along  a  line  perpendicular  to  both  planes. 

Assume  a  point  on  the  given  plane  (§  65).  Through  the 
assumed  point  draw  a  line  perpendicular  to  the  given  plane  (§  79). 
On  this  line,  measure  off  the  given  distance  from  the  starting 
point  (§  30).  Through  this  last  point  pass  a  plane  parallel  to 
the  given  plane  (§  74). 

84.  To  Determine  the  Distance  from  a  Point  to  a  Plane.  The 
required  distance  is  measured  along  a  line  through  the  point  and 
perpendicular  to  the  plane. 

First  Method.  One  method,  which  employs  no  new  operations, 
is  as  follows.  (1)  Pass  a  line  through  the  point  and  perpen- 
dicular to  the  plane  (§  79).  (2)  Find  where  this  line  pierces 
the  plane  (§  78).     (3)  Find  the  true  length  of  the  line  between 

1  Method  of  study.  The  problems  of  §§  83-88  are  quite  typical  of  many 
relatively  complex  problems,  in  that  they  involve  a  number  of  steps  in 
analysis  and  solution,  each  one  of  which,  in  itself,  is  quite  simple.  If  the 
student  allows  himself  to  become  confused,  or  if  he  does  not  thoroughly 
understand  each  of  the  underlying  problems,  he  is  lost.  On  the  other 
hand,  if  the  foundation  work  has  been  done  thoroughly,  and  if  attention 
is  centered  on  one  step  at  a  time,  both  the  analysis  and  the  construction 
may  be  built  up  quite  easily. 

In  working  up  such  a  problem,  it  is  best  to  go  through  a  complete  analysis, 
starting  with  the  given  conditions,  erecting  the  necessary  auxiliary  lines 
and  planes,  and  performing  all  the  necessary  operations  visually.  Place 
nothing  on  paper  until  the  conditions  and  the  method  of  solution  are  seen 
distinctly.  In  this  way  the  mind  is  freed  of  the  detail  of  each  individual 
operation  and  is  left  free  to  grasp  the  important  principles  involved.  This 
having  been  done,  the  detailed  working  out  of  the  construction  can  follow, 
and  attention  can  be  centered  on  the  detail  of  each  operation  without 
confusing  the  main  issue. 


VI,  §  84]  POINTS,   LIXES,   AND  PLANES 


93 


the  point  given  and  the 
piercing  point  thus 
found  (§  24). 

Second  Method.  An- 
other method  is  shown 
in  Figs.  94  and  95. 
The  point  a  and  the 
plane  Z  being  given, 
it  is  required  to  find 
the  distance  from  a  to 
Z.  Pass  the  auxihary 
plane,  Y,  through  a 
and  perpendicular  to 
ZZ^.  This  plane  will 
contain  the  required 
perpendicular  between 
a  and  Z,  and  will  intersect  Z 


Fia.  95 


in  the  line  he.     Now  let  Y  be 
rotated  about   YY^,  into  H,  so 
as  to  show   he  and   a  in  their 
actual  relations.     This  rotation 
is  shown  in  Fig.  95.     The  point 
h  falls  at  h'^,  which  is  located  in 
Fig.  94  by  making  h^h'^ 
equal   to    h%''    and    the 
angle  c^h^h'^  equal  to  90°. 
The   point   a 
rotates  to  a'^. 
The    perpen- 
dicular a'^d'^ 
can    now   be 
drawn  and 
from  it  d^  and  d''  can  be 
located  by  projecting  back 
on  eh. 


94 


DESCRIPTIVE   GEOMETRY 


[VI,  §  85 


85.  To  Determine  the  Distance  from  a  Point  to  a  Line.  The 
required  distance  is  measured  along  a  line  passing  through  the 
given  point  and  perpendicular  to  the  given  line. 

Pass  a  plane  through  the  given  point  and  perpendicular  to  the 
given  line  (§  82).  Find  where  the  given  line  pierces  the  plane 
thus  found  (§  78).  Connect  this  point  with  the  given  point, 
and  find  the  true  length  of  the  line  thus  determined  (§  24). 

86.  To  Determine  the  Common  Perpendicular  between  Two 
Lines.     The  common  perpendicular  between  two  normal-case 

lines  is  not  easy  to  visualize  nor  to  de- 
termine (§  36).     But  it  is  easy  to  con- 
struct a  line  which  is  perpendicular  to 
a  plane  (§  79).     Moreover,  if  any  two 
lines   are  given,   as  ab   and   cd,   in 
Fig.  96,  and  if  a  plane  is  passed 
through  ab  and  parallel  to 
cd,   then   any   perpen- 
dicular    from     cd 
A      to    this    plane 
will  measure  the 
distance   between  the 
two  lines,  and  one  of  these 
perpendiculars  will  be  the  com- 
mon perpendicular  between  the  two 
lines.    These  considerations  furnish  the 
basis  for  the  following  solution. 

Referring  to  Fig.  96,  the  common 
^^"  perpendicular  between  ab  and  cd 
being  required  :  (1)  Pass  the  plane  Z  through  ab  and  parallel  to 
cd  (§  72).  The  line  ef,  parallel  to  cd  and  intersecting  ab  at  any 
point,  as  g,  was  used  as  an  auxihary  line.  (2)  Draw  the  hue  ch,. 
perpendicular  to  Z  (§  79),  and  find  where  ch  pierces  Z  (§  78), 
namely,  at  j.  In  the  same  manner  determine  dk,  the  perpen- 
dicular through  d  and  touching  Z  at  k.     (3)  By  joining  j  and  k 


VI,  §  87]  POINTS,   LINES,   AND  PLANES  95 

the  projection  of  cd  on  Z  is  obtained.  This  Hne  contains  the 
foot  of  every  perpendicular  dropped  from  cd  on  Z.  (4)  The 
point  where  jk  crosses  ab,  namely,  ???,  will  be  the  foot  of  the 
common  perpendicular  between  cd  and  ab.  (5)  Draw  mn 
parallel  to  cj,  and  intersecting  cd  at  7i.  This  line  is  the  required 
common  perpendicular.  (6)  If  the  distance  between  the  lines  is 
required,  find  the  true  length  of  the  line  mn  (§  24). 

87.  Special  Case  of  Parallel  Lines.  If  the  given  lines  in 
§  86  are  parallel,  the  solution  may  be  simplified  as  follows. 
(1)  Pass  a  plane  perpendicular  to  the  two  lines.  (2)  Find 
where  each  line  pierces  this  plane.  (3)  Join  the  points  thus 
found,  obtaining  the  common  perpendicular. 

If  the  two  lines  lie  in  the  same  plane,  and  are  not  parallel,  the 
construction  is  obviously  impossible. 

EXERCISE   SHEET   XXXIII 

1.  On  the  plane  M  (73,  ^/'^  60°  r,  M"  45°  r)  locate  a  point  a,  60 
from  P  and  7  from  H.     Draw  ab  perpendicular  to  M  and  If"  long. 

2.  Given  the  points  c  (50,  11,  6)  and  d  (38,  5,  9),  through  d  pass 
a  plane  A''  perpendicular  to  cd. 

3.  Given  the  points  e  (24,  20,  6),  /  (16,  11,  9),  and  g  (III,  12,  8,  9), 
through  g  pass  a  plane  O  perpendicular  to  ef. 

EXERCISE    SHEET    XXXIV 

1.  Draw  a  plane  R  parallel  to  the  plane  Q  (68,  Q^^  30°  I,  Q"  60°  0 
and  I"  from  Q. 

2.  Given  the  points  h  (47,  12,  2),  i  (IV,  30,  2,  8),  and  j  (36,  10,  12), 
find  the  shortest  distance  from  j  to  the  line  hi. 

3.  Given  the  plane  T  (7,  T''  60°  I,  T"  30°  I)  and  the  point  I  (15,  10, 
8),  find  the  shortest  distance  from  I  to  T. 

EXERCISE   SHEET   XXXV 

1.  Given  a  (63,  3,  10),  b  (47,  15,  3),  c  (48,  2,  2),  and  d  (41,  5,  4), 
find  the  common  perpendicular  between  ab  and  cd. 

2.  Given  e  (25,  22,  8),  /  (16,  7,  3),  and  g  (13,  24,  7),  through  g 
draw  a  line  parallel  to  ef  and  find  the  common  perpendicular  between 
this  line  and  ef. 


96 


DESCRIPTIVE   GEOMETRY 


[VI,  §  88 


88.^  The  Angle  between  Two  Planes.  The  angle  between  two 
planes  is  the  angle  cut  from  the  planes  by  another  plane  passed 
perpendicular  to  their  line  of  intersection.  In  Fig.  97,  the  plane 
Z  is  perpendicular  to  mm.  According  to  the  preceding  definition, 
the  angle  hac  is  the  angle  between  the  planes  E  and  7. 

Any  plane,  as  Z,  cuts  from  tw^o  intersecting  planes  an  acute 
and  an  obtuse  angle  (6ac  and  hoK).  It  is  usual  to  speak  of  the 
acute  angle  as  the  angle  between  the  planes.  It  should  be 
recognized  that  this  is  the  greatest  angle  that  can  be  cut  from 


Fig.  97 

these  planes  by  a  third  plane  perpendicular  to  E.     For  in- 
stance, the  plane  X  cuts  out  a  smaller  angle. 

89.  The  Angle  between  a  Line  and  a  Plane.  The  angle 
between  a  line  and  a  plane  is  the  angle  between  the  line  and  its 
projection  on  that  plane.  In  Fig.  97,  the  angle  dcj  is  the  angle 
between  dr  and  U  since  X  is  perpendicular  to  //;  and  d  is  the 
angle  between  gj  and  }'.  The  method  of  determining  such  an 
angle  in  projection  is  given  in  §  98. 

90.  The  Line  of  Greatest  Declivity  of  a  Plane.  The  line  of 
greatest  declivity  of  a  plane  A  with  respect  to  another  plane  B 
is  that  line  in  the  plane  A  w^hich  makes  the  greatest  possible 

1  Arts.  88-90  inc.  arc  introduced  merely  as  a  review  of  some  of  the 
fundamental  definitions  in  solid  geometry. 


VI,  §  911  POINTS,   LINES,  AND  PLANES 


97 


angle  with  the  plane  B.  It  is  the  line  cut  out  of  ^  by  a  plane 
perpendicular  to  both  A  and  B,  and  hence  perpendicular  to  their 
intersection.  Thus  in  Fig.  97,  ab  is  the  line  of  greatest  declivity 
of  the  plane  Y  with  respect  to  H. 

Let  it  be  required  to  draw  the  line  of  greatest  declivity  of  the 
plane  Y  with  respect  to  H  in  Fig.  98  a.  Pass  a  cutting  plane 
Z  perpendicular  to  YY^.  In  space,  these  planes  have  the  same 
relations  as  the  planes  Z  and    Y  in  Fig.  93,  and  their  line  of 


Fig.  98  a 

intersection  will  be  the  required  line  of  greatest  declivity. 
This  line,  ab,  is  determined  by  the  intersections  of  the  traces, 
as  in  §  75.  The  same  plane  Y,  is  shown  in  Fig.  98  6.  Here 
the  line  of  greatest  declivity  with  respect  to  V  is  shown. 

EXERCISE   SHEET   XXXVI 

Draw  the  Hne  of  greatest  decHvity  of  each  of  the  following  planes 
with  respect  to  H  and  also  with  respect  to  V.  Determine  the  true 
angle  that  each  plane  makes  with  H  and  also  with  V.  The  planes  are : 
X  (76,  Z'^30°r,  XM5°r),  Y  (36,  F'' 30°  ^,  FM5°  r),  and  Z  parallel 
to  HA.     ZZ^  is  12  above  and  ZZ^  is  16  below  EA. 

91.  To  Rotate  a  Point  about  the  Trace  of  the  Plane  in  Which 
it  Lies.  —  Rabattement.  A  problem  which  occurs  frequently  in 
the  making  of  w^orking  drawings  consists  in  finding  the  true  shape 
of  a  figure  whose  projections  are  known  and  whose  plane  is  given. 


98 


DESCRIPTIVE   GEOMETRY 


[VI,  §  91 


When  it  was  required  to  find  the  true  length  and  slope  of  a 
line  (§  24)  the  method  consisted  in  reducing  the  line  to  a  special 
case.  Here  also  the  most  natural  method  will  be  to  bring  the 
given  figure  parallel  to,  or  into  coincidence  with,  a  plane  of 
projection.  Place  one  of  the  acute  angles  of  a  triangle  in  the 
opened  cover  of  a  book,  one  of  its  edges  touching  the  cover 
while  the  other  rests  on  the  book.     Consider  any  three  points 


Fig.  99 

on  this  plane  as  representing  a  plane  figure,  which  it  is  required 
to  rotate  into  a  position  parallel  to,  or  into  coincidence  with, 
H.  The  purpose  of  the  operation  is  to  determine  the  true 
shape  of  the  imagined  figure;  hence  the  relative  positions  of 
the  points  must  not  be  changed  during  the  rotation.  Under 
these  conditions,  it  will  readily  be  seen  that  the  most  natural 
axis  of  rotation  is  the  edge  of  the  triangle  which  rests  on  H,  and 
which  is  the  //  trace  of  the  plane  containing  the  figure. 


VI,  §  91]  POINTS,   LINES,  AND  PLANES 


99 


ax 


In  order  to  study  the  laws  governing  such  a  rotation,  let  us 
refer  to  Fig.  99,  which  shows  the  point  a,  lying  in  the  plane  Z. 
This  point  a  is  being  rotated  about  the  trace  ZZ^  as  an  axis,  into 
H.  During  the  operation,  the  radius  of  rotation,  ay,  generates 
a  plane  perpendicular  to  ZZ''.  The  lines  ay  and  a'^y  are  equal, 
since  they  are  radii  of  the  same  circle.  Now  this  radius  ay  is 
the  h^-potenuse  of  a  right  triangle  aa^y.  It  is  this  triangle  that 
gives  the  key  to  the  whole  solu- 
tion ;  hence  its  relations  to  all 
parts  of  the  problem  should  be 
thoroughly  understood  and  com- 
pletely visualized.  It  should  be 
noticed  that  the  hypotenuse  is  a 
line  of  greatest  declivity  of  the 
given  plane  with  respect  to  H. 
Moreover,  the  base  of  this  triangle 
is  the  distance  from  the  H  projec- 
tion of  the  point  to  the  H  trace  of 
the  plane,  a^y,  while  the  altitude 
of  the  triangle  is  the  height  of 
the  point  in  space,  aa'',  which  in  turn  is  equal  to  the  distance 
from  the  V  projection  of  a  to  HA,  a^x. 

Figure  100  shows  this  operation  in  projection,  the  point 
a  being  in  the  plane  Z.  It  is  required  to  rotate  it  about  ZZ^ 
as  an  axis,  into  H.  The  line  a^c  is  drawn  perpendicular  to  ZZ^. 
Since  this  line  is  the  H  projection  of  the  path  of  the  point  during 
rotation,  the  required  point  w^ill  lie  somewhere  on  it.  In  order 
to  determine  the  distance  of  the  required  point  from  y,  lay  out 
the  right  triangle  1  at  any  convenient  location  on  the  drawing. 
Make  it  equal  to  the  right  triangle  aa^y  in  Fig.  99,  i.e.  with  its 
base  equal  to  a^y  and  its  altitude  equal  to  a^x.  The  purpose 
of  this  triangle  is  merely  to  obtain  the  length  of  the  hypotenuse, 
which  is  then  measured  off  from  y,  along  the  line  a^c.  This 
gives  the  required  point,  a'^. 


100 


DESCRIPTIVE   GE0:METRY 


n,  §  91 


After  a  few  examples  have  been  solved,  the  actual  construction 
of  such  a  triangle  as  1  may  be  omitted  altogether,  the  required 
distance  being  determined  directly  with  the  dividers. 

In  Fig.  101,  the  point  b,  lying  on 
the  plane  Z,  is  rotated  about  the 
trace  ZZ',  into  V.  The  opera- 
tion is  the  same  in  principle  as 
that  shown  in  Fig.  99. 

The   process   described    in   this 
A    article,  that  of  rotating   a  point 
about  the  trace  of  a  plane  in  which 
it  lies,  is  called  rabattement. 

92.    Special  Cases.    Many  prob- 
lems involving  rabattement  may 
be     shortened     considerably     by 
recognizing,    in    this    connection, 
the  special  properties  of  points  on  the  traces  of  the  rotating  plane. 
In  Fig.  102  the  points  a  and  6,  and  the  plane  Z  are  shown  in 
Cavalier  projection,  as  is  also  the  rabattement  of  the  line  ah 


Fig.  101 


Fig.  102 


and  that  of  the  vertical  trace  ZZ'.  The  point  a  itself  lies  in  the 
axis  of  rotation  and  therefore  will  not  move  during  the  rotation. 
It  will  be  noticed  also  that  the  rabatted  position  of  h  lies  on  a  line 


VI,  §  93] 


POINTS,   LINES,  AND  PLANES 


101 


{h^h"'),  passed  through  h^  perpendicular  to  ZZ'^,  and  that  the  dis- 
tance Zh'^  is  equal  to  Z6^'. 

It  should  be  noted  particularly  that  the  line  ah,  its  H  pro- 
jection, a'^b^,  the  rabattement,  a^h'^,  and  the  H  trace  of  the  plane, 
ZZ^,  all  pass  through  the  point  a^,  which  is  the  H  trace  of  the 
given  line. 

In  Fig.  103,  the  above  operation  is  repeated  in  projection. 
It  is  required  to  rabatte  the  special-case  point  h,  into  H.  Draw 
the  line  h^h'^  through  h^,  and  perpendicular  to  ZZ^.  Swing  an 
arc  from  Z,  with  a  radius  equal  to  Zh''.     Where  the  arc  and  line 


Fig.  103 


intersect  is  the  required  point  h'^.  The  point  just  determined 
may  be  connected  with  a^,  giving  the  rabattement  of  the  line 
ah ;   or  with  Z,  giving  the  rabattement  of  the  V  trace  of  Z. 

It  should  be  noted  that  the  rabattement  of  all  points  on  Z, 
and  within  the  first  quadrant,  will  lie  within  the  angle  Z^'^ZZ^\ 

93.  To  Find  the  True  Shape  of  a  Plane  Figure  by  Rabatte- 
ment, the  Projections  and  the  Trace  of  the  Plane  Containing  it 
Being  Given. 

I.  The  General  Method.  The  figure  may  be  rotated  into 
H,  without  distortion,  by  merely  rabatting  each  of  the  ver- 
tices separately  (§  91),  and  then  connecting  the  rabatted  points. 


102 


DESCRIPTIVE  GEOMETRY 


[VI,  §  93 


VI,  §  94]  POINTS,   LIXES,  AND  PLANES  103 

II.  An  Important  Simplification.  The  preceding  method 
may  be  simpUfied  by  the  use  of  the  principle  of  §  92.  Let 
the  triangle  cde,  lying  in  the  plane  Z,  be  given,  as  shown 
in  Figs.  104  and  105.  In  Fig.  105  let  the  side  ce  be  extended 
to  its  H  and  V  traces  {g  and  h).  Next  rabatte  the  V  trace  of 
the  plane  (ZZ^),  using  the  point  h  (§  92).  The  Hne  gW^  is 
the  rabattement  of  gh,  and  the  points  e'^  and  c'^  can  be  estab- 
lished on  it  by  perpendiculars  through  e^  and  c^.  Next,  the 
rabattement  of  d  is  found  by  connecting  c'^  with/'',  and  drawing 
the  perpendicular  through  d^.     This  completes  the  triangle. 

A  check  on  the  accuracy  of  the  construction  may  be  had  by 
extending  the  side  of  the  triangle  not  used  above  {de),  to  the 
H  trace  of  the  plane,  both  in  the  projection  and  in  the  rabatte- 
ment. These  extensions  should  meet  on  ZZ^.  Again,  the 
V  projection  of  any  one  of  the  sides  may  be  extended  to  ZZ% 
and  the  point  of  intersection  may  be  rabatted  to  some  point 
on  Zh'^ ;  the  rabattement  of  this  point  should  meet  an  extension 
of  the  rabatted  side. 

94.  To  Find  the  True  Angle  between  Two  Lines.  Find  the 
traces  of  the  plane  determined  by  the  lines  (§68).  Rabatte 
the  lines  into  F  or  F  (§  93). 

EXERCISE   SHEET   XXXVII 

1.  Given  the  point  a  at  distances  59  from  P  and  12  from  H,  and  a 
point  h  at  distances  63  from  P  and  6  from  T^;  and  given  that  both 
points  lie  in  the  plane  X  (77,  X''  30°  r,  X^  45°  r).  Rabatte  a  into  H 
and  b  into  T^. 

2.  Given  the  point  c  at  distances  40  from  P  and  4  from  H,  and 
lying  in  plane  Y  (43,  Y^  Q0°  I,  F''30°r).  Rabatte  c  into  H.  Also 
rabatte  YY^  into  H. 

3.  Given  the  plane  Z  (14,  Z''  60°  r,  Z^  45°  r),  and  four  points  e,  f, 
g,  and  h,  all  of  which  Ue  in  Z  : 

e  is  in  III,  24  from  P  and  7  from  H,  rabatte  e  into  H ; 
f  is  in  I,  7  from  P  and  5  from  V,  rabatte/  into  T^ ; 
g  is  in  II,  14  from  P  and  7  from  T^  rabatte  g  into  H ; 
A  is  in  IV,  16  from  P  and  8  from  F,  rabatte  h  into  V. 


104 


DESCRIPTIVE   GEOMETRY 


[VI,  §  94 


EXERCISE   SHEET   XXXVIII 

1.  The  triangle  abc  lies  in  the  plane  S  (72,  S''  45°  r,  S''  30°  r) ;  o  is 
54  from  P  and  8  from  i/ ;  6  is  62  from  P  and  2  from  // ;  c  is  50  from 
P  and  3  from  H.  Find  the  true  shape  of  the  triangle  by  rabatting 
into  //. 

2.  The  line  joining  the  points  d  (29,  2,  13)  and  e  (15,  9,  3)  is  inter- 
sected at/,  IV'  from  d,  by /^,  where  g  is  the  given  point  (25,  13,  5). 
Find  the  true  angles  between  these  lines  by  rabattement. 

95.  Counter-Rabattement.  The  converse  problem,  counter- 
rabaitement,  as  its  name  implies,  consists  in  finding  the  pro- 
jections of  a  point,  given  its  rabattement  and  the  traces  of  the 
plane  in  which  it  is  to  lie. 

I.  First  Method.  Counter-Rabattement  by  Use  of 
THE  Line  of  Greatest  Declivity.     Given    a"^  (Fig.  106), 

as  the  rabattement  of  some  point 
in  Z ;  to  find  its  projections  after 
it  has  been  turned  back  into  Z. 
Pass  a  plane  through  a"^  and  per- 
pendicular to  ZZ^.  It  will  cut  from 
Z  a  line  of  greatest  declivity  which 
will  contain  the  point  a  (in  space). 
The  H  projection  of  this  line  of 
greatest  declivity  is  mhi^.  The  V 
projection,  not  being  needed,  is  not 
drawn.  Between  this  line,  its  H 
projection,  and  V,  lies  a  right  tri- 
angle, iVn^m^,  on  whose  hypotenuse, 
nm,,  lies  the  point  a  in  space ;  and 
on  whose  base,  n^jnJ',  lies  a^.  A  triangle  of  this  kind,  aal'y, 
is  shown  in  Fig.  99.  Let  this  triangle  be  rotated  on  ?i"# 
as  an  axis,  into  F,  giving  ii^n^m'^  as  its  true  shape.  The 
point  a  will  be  found  on  this  line,  at  a  distance  from  m'^  equal 
to  a"''m^,  the  radius  of  rotation  in  rabattement,  as  shown 
at  a'\ 


Fig.  lOG 


VI,  §  95] 


POINTS,   LINES,   AND  PLANES 


105 


This  point  {a''')  being  established,  we  know  the  height  of  the 
point  a  in  space ;  hence  a^  will  lie  on  a  horizontal  through 
a'''.  The  distance  from  m''  to  a^,  which  Avill  lie  on  m^7i^,  as 
pointed  out  above,  will  be  equal  to  m'^a'^.  It  is  now  easy  to 
locate  a^  by  measurement.  In  the  construction  used  to  draw 
Fig.  106,  the  circular  arc  a'^a^  was  used  to  make  this  measure- 
ment. The  point  a"  can  now  be  found  by  erecting  a  perpen- 
dicular from  a^. 

I.  Second  Method.  Counter-Rabattement  by  Use  of 
AN  Auxiliary  Line  Parallel  to  H.  Given  h'^  (Fig.  107) 
as  the  rabattement  of  some  point 
in  Z;  to  find  the  projections  of  h. 
Find  the  rabattement  ZZ'^  of  ZZ' 
(§  92).  Through  h'^  draw  the  ra- 
battement of  a  line  in  Z  parallel  to 
H.  The  rabattement  will  be  paral- 
lel to  ZZ^  and  will  pass  through  h'^ 
{b'^k'^).  Since  the  point  k  (in  space) 
is  on  ZZ"",  it  will  have  its  rabatte- 
ment at  k'^  and  its  H  projection  on 
HA. 

During  counter-rotation,  any 
point,  as  k'^,  will  generate  on  ^  a 
straight   line    perpendicular  to   the 

line  ZZ^,  and  the  H  projection  of  the  point  will  be  somewhere 
on  that  line.  Draw  such  a  line  through  k'^.  Its  intersection 
with  HA  will  determine  k^,  and  a  vertical  line  drawn  from  k^ 
to  ZZ'  will  give  k\ 

Through  k^  and  /:''  draw  the  projections  of  the  auxihary  line, 
k^^^  and  Z:^p''.  The  required  projections  of  the  point  h  will 
lie  on  these  lines.  As  we  have  showm  in  the  preceding  para- 
graph for  A•^  h^  will  lie  on  a  line  through  h'^,  perpendicular 
to  ZZ^.  A  vertical  Hue  through  this  point  to  h'p''  gives  the  V 
projection  of  h. 


106 


DESCRIPTIVE  GEO:\IETRY 


[VI.  §  96 


96.  To  Find  the  Projections  of  a  Given  Figure  When  it  Lies 
in  a  Given  Plane.  This  problem  requires  merely  an  extension 
and  application  of  the  principles  of  §  95,  and  constitutes  the 
general  case  of  the  problem  stated  in  §  46. 

In  Fig.  108,  let  the  plane  Z  be  given  and  let  it  be  required  to 
find  the  projections  of  a  regular  pentagon  lying  in  Z.     First 


Fig.  108 


rabatte  the  trace  ZZ""  into  //,  using  the  point  a  (§  92).  If  the 
pentagon  is  to  lie  wholly  within  the  first  quadrant,  as  is  usual, 
its  true  shape  may  be  drawn  within  the  angle  Z^'^Z  Z^,  which 
includes  the  rabattement  of  all  points  on  Z  which  lie  in  the  first 
quadrant. 

The  exact  placing  of  the  figure  within  this  space  will  de- 
pend upon  the  position  which  it  is  expected  to  take  on  the 


VI,  §  97]  POINTS,   LINES,   AND   PLANES  107 

plane  Z,  when  it  has  been  turned  up  into  that  plane.  In 
this  case,  let  it  be  so  drawn  that  one  side  in  space  will  be  parallel 
to  //.  The  rabatted  position  of  that  side  will  then  be  parallel 
to  ZZ\  as  shown  by  r^2'^,  Fig.  108. 

It  is  now  possible  to  counter-rabatte  each  of  the  vertices, 
b}'  either  of  the  methods  of  §  95.  Usually,  however,  a  simpler 
solution  is  made  possible  by  varying  the  methods  to  suit  the 
circumstances,  and  by  dealing  with  lines  instead  of  with  points. 

Let  the  side  l'^2'^  be  extended  to  the  rabatted  trace  ZZ"'^. 
The  point  of  intersection,  6'^^,  is  the  rabattement  of  the  V  trace 
of  the  line.  The  counter-rabattement  of  the  line  now  follows 
the  second  method  of  §  95.  Thus  P^''  and  1*2''  can  be  estab- 
lished. 

Next  let  the  side  2'^3'''  be  extended  to  c^  (its  H  trace). 
The  H  projection  of  2-3  passes  through  (^  and  2^  which  have 
been  found.  These  two  points  establish  the  line  2-3.  The 
point  3^"  lies  on  this  line  and  also  on  a  perpendicular  to  ZZ^, 
drawn  through  3'^.  The  V  projection  of  2-3  is  found  by  drawing 
a  line  through  c^  and  2^  and  erecting  a  perpendicular  from 
3^".  In  a  similar  manner  the  other  sides  can  be  found.  In  the 
figure,  the  auxiliary  lines  drawn  to  determine  the  sides  4-5 
and  5-1  are  omitted  for  the  sake  of  clearness. 

97.  To  Construct  the  Projections  of  a  Plane  Solid,  One  Face 
of  Which  is  in  Contact  with  a  Given  Plane.  This  problem  is 
the  general  case  of  the  one  of  which  a  special  case  was  stated  in 
§  53.  The  analysis  is  quite  simple,  consisting  principally  in 
an  application  of  the  method  of  §  96.  First,  the  base  is  counter- 
rabatted  upon  the  given  plane  (§  96).  Next  a  side  (in  the  case 
of  right  prisms),  or  the  altitude  (in  the  case  of  pyramids),  is 
drawn,  perpendicular  to  the  plane  (§  79).  The  proper  distance 
is  then  marked  off  on  this  line,  thus  fixing  the  apex  (in  the  case 
of  a  pyramid),  or  a  point  on  the  top  face  (in  the  case  of  a  prism) 
(§  30).  These  points  are  then  connected  with  the  base  pre- 
viously determined. 


108 


DESCRIPTIVE   GE0:METRY 


m,  §  97 


In  Fig.  109,  let  the  plane  A'  be  given,  and  let  it  be  required 
to  construct  the  projections  of  an  equilateral  triangular  prism 
of  given  height,  with  one  triangular  face  in  contact  with  A".  The 
plane  is  rabatted  into  H,  using  the  point  a,  on  its  T  trace,  and 
giving  the  line  A'A^'"'''.     AVithin  the  area  thus  defined,  the  true 


Fig.  109 


shape  of  the  base  {c'^h'^d'^)  is  arranged  in  the  desired  position, 
and  then  counter-rabatted  upon   A^  (§  96). 

Next,  one  of  the  vertices,  h,  is  selected,  and  the  projections 
of  a  line  of  indefinite  length  perpendicular  to  X  are  drawn 
through  the  projections  of  the  point,  to  represent  one  of  the 
edges  of  the  prism  {hH:"^  and  h^k^)  (§  79).  On  this  line  the 
given  height  is  marked  off  {hm)  (§  30). 

The  point  m  thus  found  is  one  of  the  vertices  of  the  top 


VI,  §  98]  POINTS,   LINES,   AND  PLANES  109 

face  of  the  prism.  Lines  through  c  and  d,  parallel  to  hm,  and 
other  lines  through  m  and  parallel  to  he  and  hd,  will  complete 
the  required  projections. 

EXERCISE   SHEET   XXXIX 

\.  Given  the  plane  Z  (75,  Z^2>0°r,  Z''4o°r).  Construct  the  pro- 
jections of  a  regular  hexagon  inscribed  in  a  2"  circle  and  lying  in  Z 
with  one  corner  touching  H. 

2.  Given  the  plane  Y  (26,  Y'' 30°  I,  F«60°r).  Construct  the  pro- 
jections of  a  5  pointed  star  inscribed  in  a  2"  circle  and  lying  in  Y. 
Let  the  center  of  the  circle  be  Ih"  from  YY""  and  YY^.  Start  by  ra- 
batting  the  H  trace  toward  the  right  into  V. 

EXERCISE    SHEET    XL 

Take  HA  2\"  below  the  top  border  hne.  Given  the  plane  Z  (59, 
Z^  45°  r,  Z^  30°  r),  and  using  a  scale  of  |"  =  1",  draw  the  projections 
of  a  hollow  brick,  2"  X  4"  X  8",  resting  with  one  of  its  4"  X  8"  faces 
in  contact  with  Z. 

Let  one  of  the  8''  edges  of  this  face  make  an  angle  of  30°  with  ZZ^. 
Keep  the  entire  soHd  within  the  first  quadrant. 

98.  Angle  between  a  Line  and  a  Plane.  In  order  to  determine 
the  angle  between  a  given  line  and  a  given  plane  (§  89),  first 
find  the  projection  of  the  given  line  on  the  given  plane.  This 
may  be  done  by  selecting  any  point  on  the  line,  drawing  a  per- 
pendicular from  this  point  to  the  plane  (§  79),  finding  where 
this  perpendicular  pierces  the  plane  (§  78),  and  connecting  the 
point  thus  found  with  the  trace  of  the  given  line  on  the  given 
plane. 

The  angle  thus  found  will  be  the  projeciion  of  the  required 
angle.  The  true  size  of  the  angle  may  be  found  by  the 
method  of  §  94. 

EXERCISE   SHEET   XLI 

1.  Given  the  plane  Z  (79,  Z^  30°  r,  Z^  45°  r),  and  the  points  a  (75, 
13,  16)  and  6  (67,  8,  11).     Find  the  angle  between  ah  and  Z. 

2.  Given  the  plane  Y  (19,  Y^  45°  r,  F"  60°  D,  and  the  points  c  (28, 
8,  15)  and  d  (16,  8,  8).     Find  the  angle  between  cd  and  Y. 


no 


DESCRIPTIVE   GEOMETRY 


[VI,  §  99 


99.  To  Draw  the  Traces  of  a  Plane,  Given  One  Trace  and 
the  Angle  of  Inclination  to  H  or  V. 

I.  First  Method.  In  Fig.  110,  let  the  trace  ZZ^  be  given 
and  let  the  required  angle  of  inclination  to  H  be  60°.  It  is 
required  to  determine  the  V  trace,  ZZ^ 

Imagine  a  cutting  plane,  7,  to  have  been  passed  perpen- 
dicular to  ZZ^,  at  any  convenient  point.  This  plane  will  cut 
from  Z  a  right  triangle,  the  base  of  which  is    YaJ",  the  acute 


angle  at  a  being  the  true  angle  between  Z  and  H,  or  60°.  Im- 
agine this  triangle  to  have  been  rotated  into  F,  on  Y  Y''  as  an 
axis,  a^  moving  to  a'^.  It  will  now  be  seen  in  its  true  size,  the 
hypotenuse  making  an  angle  of  60°  with  HA.  Drawing  the 
hypotenuse,  the  other  apex,  h,  is  located  at  h  .  This  is  a  point 
on  the  V  trace  of  Z,  and  the  required  trace  is  now  determined 
by  the  points  Z  and  fe". 

[Note.  The  alternative  solution  given  below  is  perhaps  more 
difficult  to  visuahze  than  that  just  given.  It  is,  however,  an  excellent 
introduction  to  §  100.] 


VI,  §  100]         POINTS,   LINES,   AND  PLANES 


111 


II.  Second  Method.  In  Fig.  Ill,  let  the  trace  ZZ^ 
be  given  and  let  the  required  angle  of  inclination  to  V  be  45°. 
It  is  required  to  determine  the  V  trace  ZZ^. 

Imagine  a  cutting  plane  Y  to  have  been  passed  perpendicular 
to  the  (unknown)  V  trace  of  Z,  at  any  convenient  point.  The 
V  trace  cannot  be  drawn  now,  but  the  B.  trace  of  such  a  plane 
will  be  perpendicular  to  II A.  Draw  YY^  to  represent  this 
auxiliary  trace.  If  the  right  triangle  cut  from  Z  by  this  auxiliary 
plane  Y  is  now  imagined  to  have  been  revolved  into  H  about 
Yh^  as  an  axis,  the  hypotenuse  will  pass  through  h^  and  will 


Fig.  Ill 


make  an  angle  of  45°  with  EA.  Now,  if  the  triangle  thus  found 
is  revolved  backward  on  the  same  axis,  the  point  a'^  will  trace 
the  arc  shown  on  F.  The  required  V  trace,  ZZ'',  will  be  tangent 
to  this  arc. 

[Note.  Compare  the  two  solutions  above  and  see  how  readily 
(6)  may  be  evolved  from  (a)  by  working  backward.] 

100.  To  Draw  the  Traces  of  a  Plane  Making  Given  Angles 
(a  and  ^3)  with  H  and  V.  In  this  problem  an  auxiliary  sphere 
with  its  center  on  HA  may  be  used  to  advantage.  Imagine 
such  a  sphere  and  the  required  plane  passed  tangent  to  it. 
Now  any  cutting  plane  passed  through  the  center  of  the  sphere 
perpendicular  to  the  E.  trace  of  the  given  plane  will  cut  a  great 
circle  from  the  sphere  and  a  line  of  greatest  declivity  from  the 
plane.     The  line  and  the  circle  will  be  tangent.  ^ 


112 


DESCRIPTIVE   GEOMETRY 


[VI,  §  100 


In  Fig.  112,  let  adc  be  the  plan  and  ahc  be  the  elevation  of 
that  part  of  the  auxiliary  sphere  which  lies  in  the  first  quadrant. 
Any  cutting  plane  passed  as  described  above  will  be  shown 
by  a  F  trace  perpendicular  to  HA  passing  through  0,  as  0''0''. 
Imagine  the  quarter  circle  and  its  tangent,  which  are  cut  by 
this  plane  from  the  quarter  sphere  and  the  required  plane, 


Fig.   112 


respectively,  to  have  been  revolved  Into  V  on  0^0'  as  an  axis. 
The  quarter  circle  will  coincide  with  cb,  and  the  tangent,  being 
a  line  of  greatest  declivity  of  the  required  plane,  will  make 
the  given  angle  a  with  HA,  since  It  here  appears  at  Its  true 
incHnation.  Draw  such  a  line,  me^.  Its  Intersection  with  0''0\ 
e^  establishes  one  point  on  the  required  V  trace.  In  a  similar 
manner,/'^  is  established  as  one  point  on  the  required  H  trace. 


VI,  §  101]         POINTS,   LINES,   AND  PLANES  113 

If  now  the  triangle  moe'-'  is  rotated  back  on  oe^  as  an  axis,  the 
point  d  will  trace  the  arc  m^m'^.  The  required  H  trace  is  tangent 
to  this  arc  (§99,  II),  and  can  be  so  drawn,  passing  through/'^. 
The  required  V  trace  is  tangent  to  w'^n'^  and  passes  through  e\ 

By  means  of  a  proof  similar  to  that  in  the  footnote  to  §  26, 
it  may  be  proved  that  the  sum  of  the  angles  which  a  plane  makes 
with  H  and  V  cannot  be  less  than  90°,  nor  more  than  180°. 

[Note.  It  is  worth  while  to  compare  the  general  method  used  in 
above  problem  with  that  of  §  29.] 

EXERCISE    SHEET    XLII 

1.  The  plane  Z  cuts  HA  at  the  point  (77,  0,  0).  The  trace  ZZ" 
makes  an  angle  of  45°  r  with  HA,  and  Z  makes  an  angle  of  75°  with 
H.     Find  ZZ^. 

2.  The  plane  Y  cuts  HA  at  (57,  0,  0),  the  trace  YY''  makes  an  angle 
of  45°  r  with  HA,  and  Y  makes  an  angle  of  45°  with  H.     Find  YY^. 

3.  The  plane  X  cuts  HA  at  the  point  (25,  0,  0),  the  trace  XX^ 
makes  45°  r  with  HA,  and  .Y  makes  an  angle  of  30°  with  H.     Find  XX^. 

EXERCISE    SHEET   XLIH 
Using  auxiliary  spheres  1-|"  in  diameter  with  centers  at  the  points 
{1)  (65,  0,  0),  {2)  (40,  0,  0),  and  (5)  (14,  0,  0),  respectively,  draw  the 
traces  of  the  planes  that  make  the  following  angles  with  H  and  V. 

(1)  The  plane  R;   45°  with  H  and  60°  with  F; 

(2)  The  plane  S ;  45°  with  H  and  30°  with  F; 

(3)  The  plane  T ;  30°  with  H  and  60°  with  F. 

EXERCISE   SHEET   XLIV 

1.  Through  the  point  a  (60,  10,  5),  pass  as  many  planes  as  possible 
that  make  angles  of  45°  with  H  and  60°  with  F,  respectively. 

[Note.  When  the  first  plane  is  found,  use  an  auxiliary  cone  with 
its  apex  at  a.] 

2.  Given  the  points  c  (22,  6,  2)  and  d  (14,  2,  5) ;  through  cd  pass  as 
many  planes  as  possible,  making  angles  of  30°  with  H. 

101.    To  Determine  the  Angle  between  Two  Planes.     The 

problem  of  determining  the  angle  between  two  plane  surfaces 
(§  88)  occurs  frequently  in  detailing  structural  work.  Several 
practical  applications  are  given  in  Chapter  X. 


114 


DESCRIPTIVE  GEOMETRY 


[VI,  §  101 


In  Fig.  113,  let  the  planes  X  and  Y  be  given,  and  let  it  be 
required  to  find  the  angle  between  them.  Any  plane  passed 
perpendicular  to  ah  will  cut  from  X  and  Y  a  triangle  similar 
to  cde,  and  the  true  shape  of  such  a  triangle  will  show  the 
required  angle  between  the  given  planes,  i.e.  dee. 

Before  attempting  the  operation  in  projection,  it  will  be  well 
to  fix  in  mind  the  geometric  relations  involved.  Since  the  plane 
of  the  triangle  dee  (the  auxiliary  plane)  is  perpendicular  to  a6, 
oc  is  perpendicular  to  ah.  Also  de  (the  H  trace  of  the  auxiliary 
plane)  is  perpendicular  to  a^h  (the  projection  of  ah)  (§  79). 
Moreover,  the  true  shape  of  the  triangle  dee  can  be  established 

if  the  base,  de,  the  point  o,  and 
the  altitude  oc,  are  known. 

We  can  now  take  up  the  opera- 
tion in  projection,  as  shown  in 
Fig.  114.     Let  the  planes  X  and 
Y  be  given,  and  the  H  projec- 
tion of   the    line   of 
intersection  {a^h^)  be 
found   (§  75).     The 
V  projection  of  this 
line  will  not  be  needed ;    hence  it  is  not  drawn. 

The  H  trace  of  the  auxiliary  plane  described  above  will  be 
perpendicular  to  a^h^.  Since  it  can  be  chosen  anywhere  along 
ah,  any  line  perpendicular  to  a%^,  as  ZZ^,  will  represent  the 
trace  of  such  a  plane.  The  points  d  and  e  are  the  vertices  of  the 
triangle  which  measures  the  required  angle  between  the  planes ; 
and  the  point  o  lies  at  the  foot  of  a  perpendicular  through  the 
apex.  Compare  Fig.  113.  As  soon  as  the  length  of  this  per- 
pendicular has  been  found,  the  triangle,  and  hence  also  the  re- 
quired angle  between  the  planes,  will  be  established. 

Turning  to  Fig.  113,  the  required  perpendicular,  oc,  is  seen 
to  lie  in  the  triangle  aa%,  as  well  as  in  the  triangle  dee.  Figure 
115  shows  this  triangle  and  the  line  oc.     If  now  aa^h  is  rotated 


Fig.  113 


VI,  §  101]  POINTS,  LINES,  AND  PLANT:S 


115 


on  a^'b  as  an  axis  into  //,  the  line  oc  will  be  seen  in  its  true 
length  and  drawn  perpendicular  to  a'b  as  shown  by  oc\ 

Returning  to  Fig.  114,  draw  a'^a'^ 

^rpendicular  to  a^b^  and  make  it 

equal     to     a'a!".      The     triangle 

is    the    true    shape 

the    corresponding 

triangle  in  Figs.  113 

and  115; 

./^  and     the 

point  0  in 

the    base 

foot    of    the 

cular    oc'^, 

now  be  drawn. 

gives    the    true 

length    of    the    altitude    of    the 

triangle  dee. 

Fig.  114  With  the  known  base  de,  the 

point  o,   and   the   altitude    just 

found,  the  true  shape  of  the  triangle  dee  can  be  constructed 

{dc"^e).     This  gives  the  required  angle  fi  between  X  and  Y. 

This  whole  construction  centers  around  the  line  oc,  which 

is  common  to  the  triangles  ad'b 
and    dee.     The    first    operation 
(drawing   the   line    ZZ^)   merely 
locates  a  definite  auxiliary-  plane, 
thus  fixing  the  points  d,  c,  and  o. 
The    second    opera- 
tion   (rabatting    the 
triangle  aa^b  into  H) 
Fig.  115  gives  the  true  length 

of  oc.    The  third  operation  places  oc  in  its  true  relation  to  de,  and 
mav  be  thought  of  as  resulting  from  the  rabattement  of  dee. 


116 


DESCRIPTIVE  GEOMETRY 


[VI,  §  101 


EXERCISE   SHEET   XLV 

Find  the  angle  between  each  of  the  following  pairs  of  planes. 

1.  Z  (75,  Z^  60°  r,  Z-  45°  r)  and  Y  ijyh,  Y>^  45°  Z,  Y-  45°  /). 

2.  X  (50,  X^  60°  r,  X-  45°  r)  and  17  (30,  IP  60°  Z,  IF-  45°  l). 

3.  f/  (25,  C/''60°r,  C7«  75°  r)  and  F  (5,  FM5°  Z,  7- 75°  0-     (See 

§76.) 

EXERCISE   SHEET   XLVI 

Find  the  angle  between  each  of  the  following  pairs  of  planes. 

1.  0  (75,  O'^  90°,  O-  30°  r)  and  P  (60,  F^  45°  I,  P-  90°). 

2.  Q  (45,  Q^  90°,  Q-  90°)  and  R  (30,  7^''  60°  I,  R^  30°  Z). 

3.  S  (13,  5''  45°  r,  S^  45°  0  and  T  (3,  T^  30°  Z,  T"  30°  I), 

102.  To  Find  the  Points  in  Which  a  Line  Pierces  a  Plane 
Solid.  The  faces  of  the  solid  are  planes  determined  by  the 
edges.  Find  the  traces  of  the  plane  faces  (§68).  Then  find 
the  point  in  which  the  given  line  pierces  each  of  these  planes 

(§  78).      If    the    given 
R"  line    pierces    the    solid, 

two  of  the  points  deter- 
mined above  will  lie 
within  the  limits  of  the 
projections  of  the  solid. 
They  are  the  required 
points.  Wlien  the  given 
solid  is  so  placed  that 
the  traces  of  the  bound- 
ing lines  fall  outside  the 
limits  of  the  drawing,  it 
is  difficult  to  determine 
the  traces  of  the  bound- 
ing planes. 

Figure  116  shows  such 
a    case.      The    tetrahe- 
dron 1-4  and  the  line  ah 
are  given  ;  the  traces  of 
Fig.  116  ah  on  1-4  are  required. 


VI,  §  104]  POINTS,   LINES,  AND  PLANES  117 

Pass  the  H  projecting  plane,  R,  through  ah.  The  traces  of 
the  lines  1-4,  2-4,  3-4  on  R  are  at  once  apparent  on  plan,  at 
c^,  d!",  e^. 

The  V  projections  of  c,  d,  and  e  are  found  by  projecting  up- 
ward from  c^,  d^y  e^,  to  the  lines  P-4^  etc.  The  line  of  inter- 
section between  the  tetrahedron  and  R  is  therefore  the  broken 
line  c'rf'e^.  The  points  /  and  g,  where  ah  crosses  cde,  are  the 
required  traces  of  ah  on  the  tetrahedron. 

103.  To  Find  the  Line  of  Intersection  between  a  Plane  and 
a  Plane  Solid.  The  points  in  which  the  edges  of  the  solid  pierce 
the  plane  will  determine  the  line  of  intersection.  Hence  if  the 
piercing  point  of  each  of  the  edges  be  found  (§  78)  and  if 
these  points  are  connected  by  straight  lines,  the  required  line  is 
determined. 

Another  method  would  be  to  find  the  line  of  intersection 
between  the  planes  bounding  the  solid  and  the  given  plane  (§  75). 

Sometimes  one  of  the  above  methods  is  better,  sometimes  the 
other.  Occasionally  the  two  may  be  combined  in  the  same 
problem.  The  selection  will  depend  on  the  exact  conditions 
in  the  given  case. 

104.  To  Find  the  Line  of  Intersection  of  Two  Plane  Solids. 

[Note.  A  special  case  of  this  problem  is  discussed  in  §  54.  The 
general  case  here  analyzed  is  that  for  solids  in  any  position.  In  execu- 
tion the  method  here  given  often  becomes  quite  cumbersome.  In  such 
cases  the  slicing  method  given  in  §  149  may  be  used.] 

If  it  is  remembered  that  the  edges  and  faces  of  plane  solids 
are  lines  and  planes,  it  will  be  seen  that  this  problem  is  merely 
an  application  of  principles  already  worked  out  in  §§  75,  78, 
102,  and  103.  In  a  particular  problem,  it  is  only  necessary  to 
carry  out  these  principles.  The  traces  of  the  planes  bounding 
the  sohds  may  be  found  and  the  lines  of  intersection  of  these 
planes  worked  out,  one  at  a  time  ;  or  the  traces  of  the  edges  of 
one  solid  on  the  faces  of  the  other  may  be  found ;  or  the  two 
methods  may  be  combined. 


118  DESCRIPTIVE   GEOMETRY  [VI,  §  104 

EXERCISE   SHEET    XLVII 

1.  The  points  a  (75,  0,  6),  6  (63,  0,  6),  and  c  (69,  0,  16i)  define  the 
base  of  a  right  triangular  prism,  2"  high.  Find  the  points  (m  and  n) 
in  which  the  hne  joining  the  points  d  (62,  1,  14)  and  e  (77,  20,  7)  pierces 
this  prism. 

2.  The  points/  (48,  0,  13),  g  (45,  0,  9),  h  (40,  0,  15),  and  i  (44,  0,  20) 
define  the  base  of  a  quadrilateral  prism.  The  points  /  and  j  (45,  12,  4) 
determine  one  edge.  Find  the  points  (o  and  p)  where  the  Une  joining 
the  points  k  (51,  14,  17)  and  I  (37,  2,  3)  pierces  this  prism. 

3.  The  point  q  (14,  0,  7)  is  the  center  of  the  base  of  a  regular  tetra- 
hedron, 1*"  on  a  side  (§50).  Find  the  line  rstu,  in  which  the  plane 
Z  (30,  Z^  30°  r,  Z"  30°  r)  cuts  this  solid. 

EXERCISE   SHEET   XLVIII 

The  points  a  (30,  0,  2),  6  (30,  0,  14),  and  o  (40,  0,  8)  define  the  base 
of  a  regular  tetrahedron,  while  the  points  d  (33,  0,  14),  e  (43,  0,  11), 
and/  (41,  0,  17)  define  the  base  of  a  triangular  prism,  of  which  the  line 
eg  (36,  7,  0)  is  one  edge.  Fmd  the  line  of  intersection  between  the 
solids.     Mark  it  h-o. 

[Note.     At  this  point  §§  183-190  can  be  profitably  studied.] 


CHAPTER  VII 
CURVED   LINES 

105.  Introduction.  Though  most  of  the  hues  in  any  structure 
are  straight,  the  use  of  curves  is  so  frequently  necessary  that  the 
designer  and  draftsman  must  always  be  prepared  to  use  them 
with  confidence  and  skill.  His  problem  is  twofold :  first,  to 
select  from  among  all  the  possible  curves  the  one  best  suited 
for  the  purpose  in  hand ;  and  second,  to  make  drawings  from 
which  his  conception  may  be  executed  precisely  by  another 
person. 

Facility  in  handling  curves  can  come  only  as  the  result  of 
a  knowledge  of  the  principles  governing  the  generation  of  curves 
in  general,  and  from  practice  on  some  specific  cases.  In  the 
following  articles  no  attempt  will  be  made  to  discuss  in  detail 
a  large  number  of  curves,  nor  to  cover  any  one  of  them  com- 
pletely. 

For  reference,  the  more  useful  rules  are  given  for  the  curves 
most  frequently  encountered  in  actual  practice.  A  few  of  the 
examples  are  included  merely  to  show  the  construction  of 
curves  that  are  typical  of  certain  groups.  Many  other  curves 
that  are  used  occasionally  in  special  classes  of  work  are  to  be 
found  in  books  dealing  with  those  special  fields. 

106.  Generation.  A  point  moving  in  a  constant  direction 
generates  a  straight  line ;  but  if  the  direction  of  motion  of  the 
point  is  constantly  changing,  a  curved  line  is  generated. 

A  curve  may  be  considered  as  made  up  of  a  series  of  adjacent 
points.^  Any  two  adjacent  points  will  determine  a  minute 
straight  line,  which  is  sometimes  called  an  element  oi  the  curve. 

^  See  the  footnote  on  page  120. 
119 


120 


DESCRIPTIVE   GEOMETRY 


[VII,  §  106 


The  direction  of  this  Hne  will  give  the  instantaneous  direction 
of  the  generating  point  when  it  passes  along  the  curve  through 
this  position.  Thus,  in  Fig.  117,  the  curve  a-h  may  be  regarded 
as  made  of  elements  which,  enlarged,  are  represented  at  a'-h'. 
The  instantaneous  direction  of  the  curve  at  d  is  represented 
at  d'e',  and  is  indicated  by  the  angles  </>  and  <^'. 


Fig.  117 


A  curve  is  sometimes  defined  as  being  generated  by  a  point 
which  moves  so  that  three  consecutive  positions  of  the  generat- 
ing point  do  not  lie,  in  general,  on  a  straight  line.^ 

107.  Tangent  and  Normal.  If  any  element  of  a  curve  is 
extended  in  its  own  direction,  as  dj,  Fig.  117,  the  straight  line 

^  The  conception  of  a  curve  as  presented  in  §§  106-107,  i.e.  that  it 
consists  of  co7isecutive  points,  no  three  of  which,  in  general,  lie  on  the  same 
straight  line,  cannot  be  defended  as  rigidly  exact,  since  between  any  two 
points  there  are  an  infinite  number  of  other  points.  Moreover,  even  ad- 
mitting the  conception  as  a  possible  one,  there  would  exist  at  a  point  of 
inflexion  (Fig.  118)  three  consecutive  points  all  in  the  same  straight  line. 
Therefore  the  use  of  the  word  consecutive  may  need  further  explanation. 
Consider  a  curve  and  two  points  P  and  Q,  on  the  curve.  A  tangent  at  P 
is  said  to  contain  two  consecutive  points  of  the  curve,  understanding  thereby 
that  it  is  the  limiting  position  of  a  secant  through  P  and  Q  when  Q  approaches 
P  as  a  limit.  The  word  consecutive  is  thus  used  in  the  sense  that  Q  can  come 
as  close  as  we  please  to  P,  along  the  curve. 

The  same  general  conception,  open  to  the  same  general  objections  and 
explicable  on  the  same  general  basis,  occurs  in  §§  108,  II,  143,  158,  161, 
and  elsewhere.  It  may  be  said  in  favor  of  the  use  of  this  conception  that 
it  has  been  found  to  aid  the  student  in  visualizing  and  classifying  curved 
lines  and  surfaces,  since  the  classifications  are  consistently  based  on  the 
same  conception  and  are  made  to  depend  on  geometric  relations  that  are 
easily  visualized. 


VII,  §  108]  CURVED  LINES  121 

so  produced  is  tangent^  to  the  curve.  Such  a  Hne  evidently 
contains  two  consecutive  points  of  the  curve  and  its  direction  is 
the  same  as  the  instantaneous  direction  of  the  curve  at  the  point 
of  tangency.  Several  cases  of  tangency  are  illustrated  in  Fig. 
118.     The  word  tangent  is  derived  from  the  Latin  word  tangere 


FiQ.  118 


(to  touch),  which  expresses  quite  closely  the  fundamental  idea 
involved. 

Any  line  perpendicular  to  a  tangent,  at  the  point  of  tangency, 
is  said  to  be  normal  to  the  curve. 

108.    Classification. 

I.  Regular  and  Freehand  Curves.  When  the  generat- 
ing point  of  any  curve  changes  its  direction  in  obedience 
to  a  definite  law,  the  curve  is  called  a  regular  curve.  In 
the  usual  case,  the  controlling  law  fixes  some  relation  that 
must  be  maintained  between  the  moving  point  and  one  or  more 
fixed  points  or  lines.  Such  examples  as  the  circle,  ellipse,  etc., 
are  readily  recalled. 

When  a  curve  is  drawn  according  to  no  definitely  stated  law% 
but  only  by  the  hand  and  eye,  the  curve  is  said  to  be  a  freehand 

1  This  conception  of  a  tangent  implies  that  a  curve  be  regarded  as  made 
up  of  a  series  of  adjacent  points  as  described  in  §  106.  Fundamentally 
the  same  idea  appears  again  in  §  143. 


122  DESCRIPTIVE  GEOMETRY  [VII,  §  108 

curve.  It  is  impossible  even  to  cite  examples  from  this  class 
as  the  lack  of  any  fixed  or  definite  characteristics  precludes 
an  effective  nomenclature. 

In  general,  freehand  curves  are  more  varied  in  character  and 
more  subtle  in  form  than  are  regular  curves.  For  this  reason 
they  are  much  used  for  ornamental  work,  especially  where 
models  or  full  size  patterns  can  be  used  to  explain  them.  For 
large  work,  however,  such  as  arches,  vaulting,  etc.,  the  use  of 
regular  curves  is  to  be  preferred  because  of  the  greater  ease 
and  precision  possible  both  in  the  drawing  and  in  the  execution. 

II.  Plane  and  Space  Curves.  Any  curve,  whether  regular 
or  freehand,  which  lies  wholly  within  a  single  plane,  is  called  a 
plane  curve.  This  class  includes  all  curves  generated  by  a 
point  moving  in  a  plane  or  by  a  plane  cutting  a  curved  surface. 

A  space  curve  may  be  thought  of  as  generated  by  a  point 
which  constantly  passes  from  one  plane  to  another,  i.e.  no  four 
consecutive  positions  of  the  generating  point  lie,  in  general,  in 
the  same  plane. ^  This  class  includes  curves  generated  by 
a  point  moving  on  a  curved  sm'face  ^  or  by  the  intersection  of 
curved  surfaces." 

109.  Projections.  Plane  curves  are  projected  in  the  same 
manner  as  other  lines  (§§  14  and  15).  The  projecting  lines  of 
the  points  on  a  curve  are  elements  of  a  projecting  cylinder 
(Fig.  119).^  The  intersection  of  this  cylinder  with  a  plane  of 
projection  gives  the  projection  of  the  curve.  This  projection 
may  vary  between  the  true  shape  of  the  curve  (a'-'b^'c'')  and  a 
straight  line  (a^b^c^).  The  projections  of  any  curve  lying  in  a 
normal-case  plane,  as  def,  will  be  curved,  but  not  equal  to  the 
curve  in  space. 

Space  curves  (Fig.  120)  have  plane  curves  for  their  projections. 
No  matter  how  a  space  curve  may  be  turned,  the  projecting 

^  See  footnote,  page  120. 

2  In  certain  special  cases  plane  curves  may  be  generated  in  this  manner. 

3  The  word  cylinder  is  here  used  in  the  general  sense  as  explained  in  §  145. 


VII,  §  109] 


CURVED   LINES 


123 


cylinder  can  never  degenerate  into  a  plane  (as  with  abc,  Fig. 
119).  Hence  the  projections  of  a  space  curve  are  always 
curved.      For   the   same   reason,  these   projections  can  never 


show  the  true  shape 
of  the  curve.  The 
true  form  of  such  a 
curve  can  be  seen 
only  by  visualizing 
it  from  its  projec- 
tions, or  by  building 
a  three-dimensional 
model. 


FiQ.  120 


124  DESCRIPTIVE   GEOMETRY  [VII,  §  109 

Let  the  student  prove  that  if  any  two  curves  are  drawn  on 
H  and  V,  at  equal  distances  from  P,  these  curves  are  the  pro- 
jections of  some  space  curve. 

Tangents.  The  projections  of  tangent  Hues  will  be  tangent 
to  the  corresponding  projections  of  the  curve.  But  the  con- 
verse of  this  proposition  is  not  necessarily  true,  as  lines  which 
appear  to  be  tangent  in  projection  may  lie  in  different  planes. 

Normals.  The  projections  of  a  normal  are  not,  in  general, 
perpendicular  to  those  of  the  tangent  (§  36). 

110.  Methods  for  Drav/ing  Curves.  Curves  may  be  drawn  in 
many  different  ways.     The  following  methods  deserve  mention. 

I.  By  Continuous  Motion  of  a  Guided  Point.  The 
generation  of  a  circle  by  compasses  is  a  familiar  example. 
In  general,  this  method  calls  for  some  arrangement  of  pivoted 
or  movable  arms,  links,  wires,  or  some  similar  device  (see  Figs. 
126,  131,  132,  and  142).  Considerable  ingenuity  can  be  used 
to  devise  ways  and  means,  which  are  adapted  to  the  work 
in  hand,  and  which  are  at  the  same  time  based  on  correct 
mathematical  principles.  The  generation  of  curves  by  con- 
tinuous motion  is  of  frequent  occurrence  both  in  the  drafting- 
room  and  in  actual  construction. 

II.  By  Establishing  Points  on  the  Curve.  Several 
points  that  lie  on  the  required  curve  are  located,  and  are  con- 
nected by  a  freehand  curve.  In  skillful  hands  this  method 
is  commonly  the  quickest,  and  it  is  quite  as  satisfactory  as 
any  other. 

This  is  the  exact  counterpart  of  the  method  used  by  a 
plasterer  in  "  spotting  "  or  ''screeding  "  an  arch.  In  drafting- 
room  practice,  particularly  for  inked  work,  the  freehand  curve 
is  sometimes  abandoned  in  favor  of  a  curve  drawn  along  one  of 
the  usual  forms  of  irregular  curves,  such  as  a  French  Curve  or  a 
Ship  Curve. 

111.  By  Approximation.  It  is  sometimes  possible  to  ap- 
proximate a  complex  curve,  such  as   an  ellipse  or  a  spiral, 


VII,  §  111] 


CURVED  LINES 


125 


by  means  of  circular  arcs  (see  Figs.  136,  137,  138).  A 
curve,  especially  an  ellipse,  drawn  in  this  manner  is  usually 
unpleasant  in  appearance,  as  it  appears  flattened  at  the  points 
where  the  arcs  are  tangent.^  Therefore  such  a  curve  is  but 
little  used  for  executed  work. 

In  the  drafting-room,  however,  this  method  may  be  used 
advantageously  to  indicate  a  curve  which  is  otherwise  fully 
described.  Moreover,  by  this  method  the  use  of  the  compass- 
pen  for  inking  is  made  possible. 

111.  To  Develop  a  Curve.  The  general  idea  of  development 
as  explained  in  §  49  is  that  of  flattening  or  straightening  out 
a  complex  surface  into  a  plane. 

I.  Plane  Curve.  Following  the  general  idea  stated  above, 
a  curve  can  be  developed  by  rolling  it  out  on  a  straight 
line.  In  the  case  of  a  circle,  this  development  can  be  obtained 
by  direct  computation, 
using  the  tt  relation. 
In  the  case  of  curves 
where  no  such  relation 
is  known,  the  develop- 
ment may  be  obtained 
by  the  use  of  instru- 
ments. 

In  Fig.  121,  the  curve 
a-h  is  to  be  developed  on  the  straight  line  a-h\  Let  the 
dividers  be  set  to  any  convenient  small  distance  and 
stepped  off  on  the  curve  from  a  to  g.  Next  step  off  the 
same  distance  on  the  straight  line  a-g\  Now  by  adding  the. 
small  remainder  gh  at  g'h\  the  total  length  of  the  curve  is 
measured  along  the  straight  line.  This  process  is  also  called 
rectifying  the  curve. 


Fig.  121 


^  By  using  a  sufficient  number  of  known  points,  the  method  of  §§  120, 
127,  and  130,  IV,  can  be  made  to  give  a  very  close  approximation  of  the 
true  curve ;   but  the  construction  is  tedious. 


126 


DESCRIPTIVE   GEO:\IETRY 


[VII,  §  111 


II.  Space  Curve.  Any  space  curve  may  be  developed  into 
a  plane  curve  by  making  it  the  directrix  of  a  cone  or 
cylinder/  the  surface  of  which  is  then  developed,  along  with 
the  given  curve,  as  explained  in   §  157  and  as  illustrated  in 


a    I 

^ 

L_d_e 

^ 

^ 

x 

s 

\ 

, Z 

1 

Q^ 

b*' 

c^ 

d*- 

e*" 

z 

a   b   c  .d    e 


(3) 
Fig.  122 

Fig.  122.  The  plane  curve  thus  determined  may  then  be 
rectified  to  a  straight  line,  as  shown  in  Fig.  122,  3,  if  desired. 

112.  The  Conic  Sections.  Curves  of  the  group  known  as  the 
conic  sections  occur  frequently  in  optics,  acoustics,  mechanics, 
and  other  sciences.  If  a  cone  is  cut  by  a  plane,  the  resulting  line 
of  intersection  will  be  one  of  the  curves  of  this  group  (see  §  145) ; 

^  The  words  cone  and  cylinder  are  used  in  the  sense  explained  in  §  145. 


VII,  §  112] 


CURVED   LINES 


127 


hence  the  name  conic  sections,  or  sometimes  simply  conies. 
The  group  consists  of  the  elHpse,  the  parabola,  and  the  hyperbola 
(including  the  circle  and  straight  line  as  special  cases). 

If  the  conies  are  considered  as  being  traced  by  a  point  moving 
in  a  plane,  methods  for  their  generation  can  be  stated  that  do 
not  involve  the  use  of  an  actual  plane  cutting  an  actual  cone. 

The  principle  of  gencratioji  applying  to  all  conies  is  that  the 
generating  point  shall  move  so  that  its  distance  from  a  fixed 
point  shall  bear  a  constant  ratio  to  its  distance  from  a  fixed 


Eccentricity  of  the  curve 


Fig.   123 


hne.  The  fixed  point  is  called  the  focus,  the  fixed  line  the 
directrix,  and  the  ratio  is  called  the  eccentricity.     (See  Fig.  123.) 

Variation  in  the  shape  of  the  different  conies  comes  from 
a  variation  in  the  eccentricity.  When  the  eccentricity  is  unity, 
the  curve  is  known  as  the  parabola.  When  the  eccentricity 
is  less  than  unity  the  curve  is  an  ellipse.  When  the  eccentricity 
is  greater  than  unity  the  curve  is  an  hyperbola. 

Variation  in  size  comes  from  increasing  or  decreasing  the 
distance  between  focus  and  directrix. 


128 


DESCRIPTIVE   GEOMETRY 


[VII,  §  113 


113.  General  Construction  for  Any  Conic.  In  Fig.  124,  let 
the  focus  /,  and  directrix  0  Y,  be  given,  and  let  it  be  required 
to  construct  a  conic  having  an  eccentricity  equal  to  n/m  (ex- 
pressed at  scale  by  the  lines  n  and  m).  Set  the  proportional 
dividers  to  the  ratio  n/m.  With  this  setting  determine  the  point 
a  which  divides  fo  in  the  given  ratio.  Draw  any  line  hs  parallel 
to  OY.     Set  the  long  leg  of  the  divider  to  the   distance  oh. 


Fig.  124 


With  the  short  leg  at  this  setting  strike  an  arc  from  /  as  a  center, 
cutting  hs  at  c.  The  point  c  is  one  point  on  the  required  curve. 
Another  line,  dh,  yields  another  point,  e.  There  are  of  course 
corresponding  points  below  the  axis.  As  many  points  as  are 
needed  may  be  found  in  this  manner,  so  that  the  curve  may  be 
drawn  as  accurately  as  may  be  required. 

Since  n/m  is<l,  the  curve  ace  is  part  of  an  ellipse.  If  the 
long  leg  of  the  dividers  had  been  used  next  the  focus,  the  result- 
ing curve  jkl,  would  have  been  an  hyperbola,  with  an  eccentricity 


VII,  §  114] 


CURVED   LINES 


129 


equal  to  m/n.     If  m  is  made  equal  to  n,  the  resulting  curve, 
urst,  is  a  parabola.^ 

114.  The  Parabola.  The  parabola  is  generated  by  a  point 
which  moves  so  that  its  distance  from  a  fixed  point  is  constantly 
equal  to  its  distance  from  a  fixed  line.     (See  §§112  and  113.) 


Since  the  eccentricity  of  every  parabola  is  the  same  the  only 
chance  for  variation  in  this  curve  lies  in  varying  the  distance 
between  the  focus  and  directrix.  Figure  125  shows  the  effect 
of  such  variation.  It  will  be  seen  that  all  have  the  same  char- 
acter, but  are  drawn  at  different  scales ;  Z  being  three  times 
as  large  as  Y  and  nine  times  as  large  as  X, 

1  The  labor  involved  in  such  problems  as  the  above  may  be  greatly  lessened 
by  a  systematic  arrangement  of  the  work.  Before  starting  work,  deter- 
mine where  points  will  be  needed  frequently  and  where  they  may  be  further 
apart.  Do  all  the  work  requiring  a  given  setting  of  tools  at  one  time. 
Search  always  the  quickest  solution.  Systematic  work  becomes  increasingly 
important  as  the  problems  become  more  complex. 


130 


DESCRIPTIVE  GEOMETRY 


[VII,  §  114 


It  will  be  evident  that  as  the  focus  approaches  the  directrix, 
the  curve  approaches  the  axis ;  while  as  the  focus  recedes,  the 
curve  also  recedes,  approaching  a  line  parallel  to  the  directrix. 

115.    To  draw  the  Parabola,  given  the  Directrix  and  Focus. 

I.  By  Continuous  Motion  (§  110,  I).  Let  the  directrix, 
Fig.    126,   be  so   placed  as  to    coincide  with    the  edge  of   a 


^. 


■'k(  Pencil) 


a 


Thread ,  eaud  in 
length  to  qd. 


Axi5 


Fig.  126 

drawing  board,  cd,  and  let  the  T-square  be  fitted  with  a  thread 
akg,  as  shown.  Let  the  blade  of  the  square  be  placed  so  as  to 
coincide  with  the  axis.  The  thread  will  now  loop  from  g  through 
/,  and  back  to  a'.  Place  a  pencil  in  the  loop,  at  /,  drawing 
the  thread  taut.  Keeping  the  pencil  within  the  loop  and  in 
contact  with   the    edge   of    the   square,   slide   the  square   up- 


VII,  §  116] 


CURVED  LINES 


131 


wards.  The  upper  half  of  the  curve  fi  will  be  drawn.  The 
lower  half  may  be  drawn  in  the  same  manner,  fastening  the 
thread  at  h.^ 

II.  By  Establishing  Points  (§110,11).  This  is  a  special 
case  of  §  113,  where  the  ratio  n/m  =  l.  In  this  case  the  pro- 
portional dividers  may  be  replaced  by  the  ordinary  type  of 
instrument.^ 

116.  To  draw  the  Parabola,  given  the  Axis,  the  Vertex,  and  one 
other  Point  on  the  Curve.     In  Fig.  127,  let  c  be  the  given  vertex, 


b 

9 

7         5 

3 

1 

2/^ 

2 

/ 

1 

/    / 

A 

^ 

^  • 

4 

/ 

/ 

y 

6 

/   / 

/ 

Axis 

8 

/// 

i 

10 
C 

W 

A 

FiG.    Vll 


a  the  other  point,  and  cd,  the  axis.  Through  c  draw  a  line 
perpendicular  to  cd,  and  through  a  draw  a  line  parallel  to  cd. 
Divide  these  lines  into  the  same  number  of  equal  parts,  as  shown 
by  points  1-10.  Connect  c  with  1,  3,  5,  etc.,  and  draw  lines 
parallel  to  cd  through  2,  4,  6,  etc.  The  points  of  intersection, 
^j  />  Qi  etc.,  are  on  the  required  parabola.  This  construction 
is  based  on  the  same  idea  as  that  of  §  124,  II  (1),  for  the  ellipse, 
and  §  130,  III,  for  the  hyperbola. 

1  Practical  difficulties  that  arise  in  fastening  the  thread  and  in  getting  a 
thread  that  will  not  stretch  make  this  method  more  of  theoretical  than  of 
practical  value.  This  criticism  also  applies  to  the  methods  of  §§  124,  I  and 
130,  V. 

2  See  footnote,  page  129. 


132 


DESCRIPTIVE   GEOMETRY 


[VII,  §  117 


117.  To  find  the  Focus  of  a  given  Parabola.  In  Fig.  128, 
choose  any  point  m  on  the  curve.  Draw  vm  perpendicular  to 
the  axis.  On  this  Hne,  lay  off  np  equal  to  twice  vn.  Draw  pv, 
cutting  the  curve  at  r.  From  r  drop  a  perpendicular  to  the 
axis.     The  foot  of  this  perpendicular,  /,  is  the  required  focus. 

118.  To  draw  a 
Tangent  to  a  given 
Parabola. 

I.  At  a  Given 
Point  on  the 
Curve.  First 
method.  In  Fig. 
128,  let  a  be  the 
given  point  of 
tangency.  Draw 
ab  perpendicular  to 
the  'axis.  Lay  off 
vc  equal  to  vb.  The 
required  tangent  is 
the  line  ca. 

Second  method, 
using  the  focus. 
Let  d  be  the  re- 
quired point  of 
tangency.  The 
focus,  /,  can  be  de- 
termined by  §  117. 
Lay  off  vz  =  vf,  and 
draw  the  directrix,  xy.  Connect  d  with  the  focus,  /.  Draw 
dg  parallel  to  the  axis  and  bisect  the  angle  gdf.  This  bisector 
de  is  the  required  tangent.  Compare  with  §§  125,  I  and 
130  a. 

.  11.   Through  a  Point  Outside  the  Curve.     Let  s,  Fig. 
128,  be  the  given  point.     With  5  as    a  center  and  sf  as  a 


Fig.  128 


VII,  §  120] 


CURVED  LINES 


133 


radius,  draw  a  circle  tfw  intersecting  the  directrix  at  t  and  w. 
Perpendiculars  to  the  directrix  drawn  from  these  points  will 
intersect  the  parabola  at  q  and  u  which  are  the  required  points 
of  tangency. 

119.  To  draw  a  Normal  to  a  given  Parabola. 

First  method.  One  obvious  method  consists  in  first  drawing 
a  tangent,  then  drawing  a  perpendicular  to  the  tangent  at  the 
point  of  tangency. 

Second  method.  Another  method,  using  the  focus,  follows. 
(See  Fig.  128.)  Let  h  be  the  point  at  which  the  normal  is  to  be 
drawn.  Draw  hj  per- 
pendicular to  the  axis. 
Lay  off  jk  equal  to 
twice  vf.  The  required 
normal  passes  through 
k  and  h. 

120.  To  approximate 
a  Parabola  by  Arcs  of 
Circles.  Establish  any 
number  of  points  that 
are  known  to  lie  on  the 
parabola  (Fig.  129, 
points  Vy  1,  2,  3,  etc.), 
by  the  method  of 
§§  115,  II,  or  116. 
Draw   the   normals   at 

these  points  (§  119).  The  axis  is  a  normal  at  the  vertex. 
Prolong  these  normals  to  their  intersections  at  a,  h,  and  c. 
With  a  radius  equal  to  av  strike  the  circular  arc  v-\,  from  the 
center  a.  Move  the  center  to  h,  and  using  the  radius  h-\, 
strike  the  arc  1-2.     Continue  for  each  center.^ 


Note.  c 

f  =  focus 

X2,"L)3    etc.,  II  0X15. 

2b  II  xf .  3C  II  yf  ietc. 


Fig.  129 


1  See  footnotes  to  §§  110  and  113.  Also  note  that  the  principle  here 
used  may  be  applied  to  the  construction  of  an  ellipse  (§  127,  III)  or  of 
an  hyperbola. 


134 


DESCRIPTR^   GEOMETRY 


[VII,  §  121 


121.  The  Ellipse.  An  ellipse  is  generated  by  a  point  which 
moves  so  that  its  distance  from  a  fixed  point  bears  to  its  dis- 
tance from  a  fixed  line  a  constant  ratio  less  than  unity.  (See 
also  §  112.)  For  typical  form  and  nomenclature  for  the  ellipse 
see  Fig.  130.  The  symmetry  of  the  curve  gives  two  foci  and 
two  directrices.     For  any  point  on  the  curve,  as  d,  df/dp,  and 

V  and  v'  =  vertices. 

hj  and  ge  =  conjugate  diomeiers  .  hj  being 

parallel  to  tanqents  at  g  and  e. 


/ 

nf  =  cv. 

kf :  kl ; :  df :  dp: :  df :  dp' ; :  vf :  vo  <  1 .  =  eccentricity.. 
df  +  clf=  vf+vf'=  vv". 
df  and  df'=  focal  distances  of  d. 
Fig.  130 

df/dp'  are  equal  to  the  eccentricity  of  the  curve.  Also  cf/eo 
is  the  same  ratio. 

The  ellipse  may  also  be  defined  as  a  curve  generated  by  a 
point  which  moves  so  that  the  sum  of  its  distances  from  two 
fixed  points  remains  the  same ;  that  is  to  say,  in  the  figure, 
df-^df  =  hf+hf  =  vf+vf  =  the  major  axis. 

The  shape  of  the  ellipse  will  vary  with  the  eccentricity  and 
with  the  distance  between  focus  and  directrix.  As  the  eccen- 
tricity approaches  zero,  the  curve  approaches  a  circle  or  a  point. 


VII,  I  124] 


CURVED  LINES 


135 


As  the  eccentricity  approaches  1,  the  curve  approaches  its  axis, 
or  a  parabola. 

122.  To  construct  an  Ellipse,  given  the  Focus,  Directrix,  and 
Eccentricity.     This  is  a  special  case  of  §  113,  which  see. 

123.  To  find  the  Foci  of  an  EUipse.  From  §  121.it  will  be 
seen  that  in  Fig.  130,  nf-\-nf'  =  vv' .  Therefore  if  an  arc  is 
struck  from  n  as  a  center,  with  a  radius  equal  to  half  the  major 
axis,  the  points  /  and  f,  where  this  arc  cuts  the  axis,  will  be 
the  required  foci. 

124.  To  construct  an  Ellipse,  given  Major  and  Minor  Axes. 
I.   By  Continuous  Motion.     (1)  First  meihod.     In  Fig.  131, 

let  ah  and  vv'  be  the  axes.     By  §  123,  locate  the  foci,  /  and  /'. 


Fig.  131 


Put  pins  at  the  foci  and  loop  around  them  a  continuous  string, 
the  length  of  which  equals  2fv.  A  pencil,  p,  sliding  in  the  loop 
will  describe  the  required  ellipse.  For  large  size  rough  work, 
such  as  gardening,  this  method  is  satisfactory,  but  for  small 
or  accurate  constructions,  the  difficulty  of  keeping  the  string 
taut  but  without  stretching  is  too  great.  Compare  with  §  115, 1. 
(2)  Second  method.  A  method  which  gives  better  results 
than  the  former,  is  the  construction  by  means  of  a  trammel  ^ 

1  A  trammel  is  a  mechanical  dev'ice  for  constructing  curves  by  continuous 
motion.  The  term  is  sometimes  applied  to  beam  compasses,  though  not 
usually  applied  to  compasses  of  smaller  size. 


136 


DESCRIPTIVE   GEO^IETRY 


[VII,  §  124 


as  shown  in  Fig.  132.     Thin  strips  are  tacked  down  along  the 
axes  ab  and  vv\     A  lath  pmn  is  bored  as  shown,  p77i  being  equal 


Fig.   132 

to  ao  and  p?i  equal  to  vo,  pegs  inserted  at  w?  and  n,  and  a  pencil 
is  placed  at  p.  The  pegs  being  inserted  in  the  grooves  between 
the  strips,  the  motion  of  the  lath  will  cause  the  pencil  to  de- 
scribe an  ellipse. 

e^ -^^^-1-^^^ — ^/—^T  ,*'  11.   By     Estab- 

LisHixG  Points. 
(1)  Firsi  method. 
In  Fig.  133,  let  ab 
and  cd  be  the  given 
axes.  Construct  a 
rectangle  efgh  on 
these  diameters. 
Divide  ao  and  ae 
into  the  same  num- 
ber of  equal  parts 
as  shown.  Draw 
d  1,  d  2,  etc.,  and 
cl',  c2',etc.  The 
intersections  x,  y, 
etc.,  lie  on  the  required  curve.  This  method  resembles  that  of 
§§  116  and  130,  III.  It  applies  to  either  the  ellipse  or  the  circle. 
It  may  be  used  without  change  when  conjugate  diameters  (see 


pr  =pz. 

psll  rz. 

V 

'» 

Z'k=  t\=  ab: 

X 

^-  \k/ 

Fig.  133 

VII,  §  1251 


CURVED  LINES 


137 


Fig.  134 


Fig.  130)  are  given  instead  of  the  main  axes,  except  that  the 
rectangle  efgh  will  be  replaced  by  a  parallclo^^ram  as  in  Fig.  134. 

(2)  Second 
method.  In  I  (2) 
above,  a  piece 
of  paper,  marked 
at  points  corre- 
sponding to  m,  71, 
and  p,  may  be 
substituted  for 
the  lath.  If  the 
paper  is  placed 
so  that  m  and  n 
fall  on  the  axis  lines,  the  point  p  is  a  point  on  the  eUipse. 

(3)  Third  method.     In  Fig.  135,  let  db  and  eg  be  the  major 

and  minor  axes. 
Draw  the  inscribed 
and  circumscribed 
circles,  and  the 
radii  oj,  ok,  etc. 
Draw  verticals  j/, 
kk\  etc.,  and  hori- 
zontals   J"ff    k"k'  y 

etc.  The  intersec- 
tions /,  k',  etc., 
are  on  the  required 
ellipse. 

125.   To  draw  a 
Tangent  to  an  El- 
FiG.  135  lipse. 

I.  At  a  Point  on  the  Curve.  (1)  First  method.  Sup- 
pose the  ellipse  drawn  as  in  Fig.  135.  Let  k'  be  the  given  point 
of  tangency.  Draw  mn  and  kl  tangent  to  the  inner  and  outer 
circles  at  k"  and  k.     The  line  ml  is  the  required  tangent. 


138  DESCRIPTIVE   GEOMETRY  [VII,  §  125 

(2)  Second  virfhod.  In  Fig.  133  another  construction  is 
shown.  Let  it  be  required  to  draw  a  tangent  to  the  elHpse 
at  p.  Connect  p  with  the  foci,  z  and  z'.  Extend  z'p  to  r  making 
pr  equal  to  pz.  Bisect  the  angle  rpz.  The  bisector  pm  is 
tangent  to  the  elhpse.  The  center  point  of  zr  determines  the 
bisector.     Compare  with  §§118  and  130  a. 

II.  Through  a  Point  Outside  the  Curve.  In  Fig.  133,  let 
q  be  the  given  point.  With  g  as  a  center  and  qz  as  a  radius, 
draw  the  circle  zjk.  "With  z'  as  a  center  and  a  radius  equal  to 
ah,  draw  an  arc  intersecting  the  circle  zjk  at  j  and  k.  Draw 
z'j  and  z'k,  intersecting  the  ellipse  at  u  and  v,  which  are  the 
required  points  of  tangency. 

126.  To  draw  a  Normal  to  an  Ellipse.  Using  the  construction 
described  in  §  125,  I,  (2),  and  shown  in  Fig.  133,  a  line  ps 
through  p  parallel  to  rz  will  be  normal  to  the  ellipse. 

127.  To  approximate  an  Ellipse  by  Arcs  of  a  Circle.  IMany 
rules  for  approximating  to  an  ellipse  have  been  proposed.  (See 
§  110,  III.)  Of  course  any  such  rule  will  give  a  curve  which 
varies  more  or  less  from  a  true  ellipse.  Some  of  the  following 
methods  give  a  closer  approximation  than  others.  Some  are 
useful  only  for  curves  that  have  a  certain  ratio  of  major  to 
minor  axis.  Some  are  much  more  tedious  than  others.  In 
general  the  greater  the  number  of  centers  used,  the  closer  the 
approximation,  and  the  greater  the  labor  involved. 

I.  Four  Centers  (Fig.  136).  Given  the  axes  cm'  and 
bb'.  Lay  off  ac^oh,  and  oc'  =oc.  Bisect  cc'  at  d,  lay  off  cc^cd, 
and  of  =  oe.  The  points  /'  and  e'  are  symmetric  to  /  and  e.  The 
four  points  serve  as  centers  from  which  arcs  can  be  struck, 
which  will  be  tangent  to  one  another,  and  which  will  approximate 
an  ellipse.  This  method  is  not  useful  for  narrow^  curves,  but 
it  may  be  executed  rapidly.  When  one  half  of  this  curve  is 
used  for  an  arch,  the  arch  is  called  a  three-centered  arch. 

II.  Eight  Centers  (Fig.  137).  Given  the  axis  ab  and  the 
semi-axis  oc.     Draw  the  rectangle  CLefb.     Connect  a  with  c,  and 


VII,  §  127] 


CURVED  LINES 


139 


draw  ep  perpendicular  to  ac.  Lay  off  or  =  oc,  and  on  ar  as  a 
diameter,  construct  the  semicircle  ajlr.  Draw  the  radius  jg, 
perpendicular  to  ab. 
Lay  off  ok  =  h  j,  and 
with  pk  as  a  radius, 
and  with  pas  a  center, 
swing  the  arc  mkm'. 
Extend  oc  to  I.  Using 
a  and  h  as  centers, 
and  a  radius  equal  to 
ol,  strike  arcs  cutting 
mkvi'  at  m  and  m'. 
The  required  centers 
are  at  n,  m,  p,  m',  and 
n\  This  method  is  applicable  to  ellipses  of  nearly  all  propor- 
tions.    An  arch  following  this  curve  is  called  a  five-centered  arch. 

-41 


\\f/ 


Fig.  137   i 


140 


DESCRIPTIVE  GEOMETRY 


[VII,  §  127 


III.  Numerous  Centers  (Fig.  138).  The  major  and  minor 
axes  are  given.  The  first  step  is  to  locate  the  foci,  /  and  /' 
(§  123).  Next  locate  a  number  of  points,  a,  b,  c,  etc.,  which  are 
known  to  lie  on  a  true  ellipse  (§  124,  II).  At  the  points  thus 
located,  draw  normals  to  the  ellipse,  ah,  hi,  etc.  (§  126).  The 
points  1,  2,  3,  etc.,  where  these  normals  intersect,  may  be  used 
as  centers  for  striking  the  curve.  The  construction  for  any 
normal,  as  cj,  is  made  by  making  eg  equal  to  cf,  and  drawing 


g~\'-- 


cj  parallel  to  gf.  If  g  and  /  are  known,  the  line  cj  can  be  drawn 
parallel  to  gf  without  actually  drawing  the  line  gf,  thus  simplify- 
ing the  drawing  and  shortening  the  work. 

128.  To  draw  the  Projections  of  a  Circle  inclined  in  any 
Manner  to  H  and  V.  The  key  to  this  problem  lies  in  the  fact 
that,  in  whatever  position  a  circle  may  be  placed  with  respect 
to  H  (or  V),  some  one  of  its  diameters  will  be  parallel  to  H 
(or  V),  and  hence  will  be  projected  on  H  (or  V)  in  its  true 
length.  This  being  the  greatest  possible  projection  of  any 
diameter,  it  must  form  the  major  axis  of  the  ellipse  which  is 


VII,  §  128] 


CURVED   LINES 


141 


the  required  projection  of  the  given  circle.  By  the  same  token, 
that  diameter  of  the  circle  which  is  a  line  of  greatest  declivity 
of  the  plane  in  which  the  circle  lies  will  have  the  least  possible 
projection  of  any  diameter ;  and  hence  will  be  the  minor  axis 
of  the  projection. 


Fig.  139 


In  Fig.  139,  let  it  be  required  to  draw  the  projections  of  a 
circle  lying  in  the  plane  X,  with  its  center  at  o.  Rabatte  the 
V  trace,  X  A^^',  and  the  center  o,  into  H  (§§  91  and  92).  Draw 
a  circle  of  the  required  size  with  its  center  at  o'^.  Draw  the 
diameters  which  are  parallel  and  perpendicular  to  X  X^.    When 


142  DESCRIPTIVE   GEO:\IETRY  [VII,  §  129 

these  lines  are  rotated  back  into  X,  they  will  give  the  major  and 
minor  axes  of  the  H  projection,  since  one  will  be  parallel  to  //, 
while  the  other  is  a  line  of  greatest  declivity  of  the  plane.  The 
lines  a^b^  and  c^d^  are  the  counter-rabattements  as  seen  on  H 
(§  95).  The  ellipse  may  now  be  drawn  by  any  of  the  methods  of 
§  124.  The  F  projection  of  these  axes  being  found  (§  95),  they 
are  conjugate  diameters  of  the  required  ellipse  (§  124,  II,  (1)). 

Let  the  student  determine  the  diameters  of  the  rabatted  circle 
which,  by  counter-rabattement,  will  give  the  major  and  minor 
axes  of  the  V  projection. 

129.  The  Hyperbola.  An  hyperbola  is  generated  by  a  point 
which  moves  so  that  its  distance  from  a  fixed  point  bears  to  its 
distance  from  a  fixed  line  a  constant  ratio,  greater  than  unity. 
(See  also  §  112.)  As  in  the  case  of  the  ellipse,  there  are  two 
foci,  two  directrices,  and  two  axes,  but  the  foci  lie  outside  the 
directrices,  and  the  curve  is  composed  of  two  symmetrical  parts, 
branching  off  to  infinity. 

An  important  property  of  the  hyperbola  is  that  the  difference 
between  the  focal  distances  of  any  point  on  the  curve  is  always 
the  same,  and  is  equal  to  the  distance  between  the  vertices. 
This  property  is  closely  parallel  to  a  similar  property  of  the 
ellipse  stated  in  §  121.  The  hyperbola  is  met  with  so  infrequently 
in  practice  that  only  a  few  constructions  will  be  given. 

130.  To  Construct  an  Hyperbola. 

I.  The  General  Construction.  The  method  described  in 
§  113  applies  to  the  hyperbola  as  well  as  to  all  other  conic 
sections. 

II.  A  Construction  Based  on  the  Difference  of  Focal 
Distances.  This  method  is  shown  in  Fig.  140.  Given  the 
foci  F  and  F'  and  the  vertices  v  and  v\  From  F  and  F'  as 
centers,  with  radii  greater  than  Fv,  swing  arcs  aa  and  a'a\  Add 
to  the  radius  just  used  an  amount  equal  to  vv\  and  swing  arcs 
bb  and  b'b'  from  the  centers  F'  and  F.  The  intersections  c 
and  cf  are  points  on  the  hyperbola. 


VII,  §  130  a] 


CURVED  LINES 


143 


III.  Another  method,  similar  to  that  of  §§  116  and  124,  II,  1 
is  shown  by  the  dotted  Hnes  in  Fig.  140. 

IV.  The  hyperbola  may  be  approximated  by  circular  arcs, 
following  the  method  used  in  §§  120  and  127,  III. 


3         2  I 


Fig.  140 


V.  A  method  similar  to  that  of  §§  115, 1,  and  124, 1,  may  be 
devised  for  drawing  an  hyperbola  by  continuous  motion. 

130  a.  Tangent  and  Normal  to  the  Hyperbola.  The  tangent 
to  an  hyperbola  bisects  the  angle  between  the  focal  radii  of  any 
point. 

Thus,  in  Fig.  140,  the  angle  F'pm  being  made  equal  to  the 
angle  mpF,  the  line  vip  is  tangent  to  the  hyperbola  at  p. 

To  draw  a  normal,  produce  Fp  to  r  making  pr  =  pF'.  Draw  r  F', 
and  bisect  it  at  n.  The  line  pn  is  the  required  normal.  (Com- 
pare with  §§  118,  II;   125,  I,  (2)  and  126.) 


144 


DESCRIPTIVE   GEOMETRY 


[VII,  §  131 


131.  The  Sinusoid.  The  curve  shown  in  Fig.  141,  is  called 
a  sinusoid  or  a  sine  curve.  In  this  curve  the  ordinates  (vertical 
distances  from  the  horizontal  axis)  are  proportional  to  the  sines 
of  the  abscissae  (horizontal  distances  from  the  origin). 

This  curve  is  introduced  to  call  attention  to  a  curve  of  contra- 
flexure,  i.e.  one  in  which  the  direction  of  curvature  changes 
periodically.  It  will  be  noticed  that  as  the  curve  approaches 
the  axis,  its  curvature  becomes  quite  flat  while  the  sharpest 
point  is  farthest  from  the  axis.  At  the  axis,  the  upper  and  lower 
curves  have  a  common  tangent.  WTiere  brackets,  moldings, 
etc.,  are  designed  with  outlines  composed  of  reverse  curves, 
even  when  freehand  curves  are  used,  the  principle  of  flattening 
the  parts  as  they  approach  their  common  tangent,  is  adhered  to. 


In  the  figure,  the  distances  0-1,  1-2,  2-3,  etc.,  are  equal,  and 
each  is  equal  to  1/24  of  the  circumference  of  the  circle  from  which 
the  ordinates  are  derived.  Flatter  or  steeper  sine  curves  could 
be  drawn  by  increasing  or  decreasing  the  ratio  between  0-1,  1-2, 
etc.,  and  0-1',  l'-2',  etc.  All  such  sine  curves  belong  to  the 
general  family  of  harmonic  curves. 

132.  Link  Motion  Curves.  —  The  Lemnlscate.  Figure  142 
shows  three  links,  AB,  CD,  and  BDE,  pivoted  at  A,  B,  C,  and 
D.  The  pivots  A  and  C  are  stationary,  while  B  and  D  are 
free  to  move.  Obviously  B  can  move  only  in  a  circular  path 
about  ^  as  a  center.  Also  D  is  confined  to  a  circular  path 
about  C.  Moreover,  B  and  D  must  always  be  equally  distant 
from  one  another,  the  constant  distance  being  fixed  by  BD. 
Thus  the  points  B  and  D  have  a  simple  circular  motion.     But 


VII,  §  132] 


CURVED   LINES 


145 


the  motion  of  any  other  point  on  BDE  is  a  compound  motion. 
To  trace  the  path  of  E,  let  the  circular  paths  of  B  and  D  be 
drawn,  giving  the  circles  a-e-g-v  and  g'h'q' .  Divide  a-v  into 
any  number  of  parts.  In  the  figure  sixteen  equal  parts  were 
used,  with  a  few  intermediate  points  for  accuracy.  Assume 
that  D  is  moved  toward  /,  passing  successively  through  the 


EO  =  DB-CO 
AB  =  AC 
CD'A&::  2:3 


Fig.  142 


other  points.  Meanwhile  B  is  traveling  along  q'h'g' ;  at  the 
distance  BD  from  the  points  /,  g,  h,  etc.,  successively.  The 
positions  of  B  which  correspond  to/,  g,  h,  etc.,  may  be  found  by 
striking  arcs  from  these  points  with  a  radius  of  BD.  The 
intersections  of  these  arcs  with  g'b'q'  give  the  points  /,  g',  etc., 
which  mark  successive  positions  of  B.  Through  corresponding 
points  (/  and  f,  g  and  g',  etc.)  draw  Hues  equal  to  BE.  The 
extremities  of  these  lines  give  the  points  /i,  gi,  etc.,  which  are  on 
the  required  curve.  It  will  be  noted  that  the  path  of  B  is 
confined  to  somewhat  less  than  half  of  the  circle  g'b'q'. 


146 


DESCRIPTIVE   GEOiVIETRY 


[VII,  §  132 


The  particular  curve  shown  in  Fig.  142  is  a  lemniscate  because  of 
the  proportion  between  the  Unks  indicated  in  the  figure.  Interest- 
ing variations  may  be  found  by  varying  these  proportions  or  by 
tracing  the  path  of  some  other  point  on  the  free  Hnk.  The  prin- 
ciple of  curvature  at  points  of  contraflexure,  noted  in  connection 
with  the  sinusoid,  will  be  seen  to  hold  good  for  this  curve  as 
well.  Similar  problems  arise  in  connection  with  hinging  of 
certain  types  of  doors,  casements,  etc. 

133.    The   Cycloid.     The  cycloid  is  a  curve  traced  by  any 
point  on  the  circumference  of  a  circle  which  is  made  to  roll  on 
a  straight  line,  as  in  the  case  of  a  wheel  rolling  on  a  rail. 
7  a* 


JTI     12 


Fig.   143 

In  Fig.  143  the  circle  1-12  rolls  on  the  base  vi-n.  Let  it  be 
required  to  trace  the  path  of  the  point  a.  After  one  complete 
revolution  of  the  circle,  the  point  a  will  again  be  in  contact  with 
the  base  line,  at  a  distance  from  its  original  position  equal  to 
27rr.  This  locates  the  point  a^^.  The  center  o  moves  forward 
on  a  level  Hne  oo^^.  As  the  points  2,  3,  4,  etc.,  come  in  contact 
with  the  base,  at  2^  3',  etc.,  the  center  moves  to  o^,  o^  etc., 
directly  above.  When  2  is  at  2',  the  lowering  of  2  will  have 
resulted  in  an  equal  raising  of  a.  At  the  same  instant  the 
distance  of  a  from  o  is  equal  to  the  radius  of  the  circle.  Thus, 
by  swinging  arcs  of  radius  equal  to  that  of  the  generating  circle, 
from  the  centers  o\  cr,  o^,  etc.,  cutting  the  lines  of  level,  2-12, 
3-11,  etc.,  the  points,  a},  o?,  a^,  etc.,  on  the  cycloid  are  estab- 
lished. 


VII,  §  134] 


CURVED   LINES 


147 


The  cycloid  Is  a  special  case  of  the  family  of  curves  called 
trochoids.  This  family  includes  all  curves  generated  by  a 
point  on  a  circle  which  rolls  in  contact  with  another  circle. 
In  the  case  of  the  cycloid  the  base  circle  is  of  infinite  radius. 
If  the  radius  of  the  base  circle  is  finite,  and  the  rolling  circle  is 
outside  of  the  base  circle,  the  curve  generated  is  an  epicycloid. 
If  the  rolling  circle  is  within  the  base,  the  curve  is  an  hypo- 
cycloid.  Various  other  trochoids  result  from  tracing  the  paths 
of  points  within  or  outside  the  circumference  of  the  rolling 
circle. 

134.  The  Involute  of  a  Circle.  This  curve  is  generated  by 
unwrapping  a  thread  from  a  circular  drum.     In  Fig.  144  let 


Fig.  144 


the  circle  1-12  be  a  plan  of  the  cylinder  from  which  the  thread 
is  to  be  unwound,  starting  at  1.  ^^^len  one  complete  turn  has 
been  made,  the  thread  will  be  tangent  to  the  circle  at  1  and  the 
end  of  the  string  will  be  at  a  distance  from  1  equal  to  the  cir- 
cumference of  the  circle,  as  shown  by  the  line  1-m.  At  other 
points  on  the  circumference,  the  thread  will  always  be  tangent 
to  the  circle,  the  distance  from  the  point  of  tangency  being 
equal  to  a  proportionate  part  of  the  circumference. 


148 


DESCRIPTIVE   GEOMETRY 


[VII,  §  134 


Of  course  this  curve  may  consist  of  as  many  turns  as  may 
be  desired,  the  general  shape  being  that  of  a  spiral.  The  circle 
from  which  the  curve  is  evolved  is  called  the  evolute.  Any 
plane  figure  may  be  used  as  an  evolute  to  form  a  spiral  curve, 
which  is  called  its  involute. 

135.  The  Ionic  Volute.  This  curve,  as  commonly  con- 
structed, is  the  involute  of  a  diminishing  square.     The  con- 


FiG.  145  (After  Vignola  ) 

struction  is  completely  shown  in  Fig.  145.  It  is  composed  of 
mutually  tangent  circular  arcs.  The  centers  for  these  arcs 
are  located  in  the  order  shown  on  the  large  scale  insert  in  the 
figure,  the  small  circles  indicating  the  centers  for  the  outer 
curve,  while  the  short  cross  lines  indicate  the  corresponding 
centers  for  the  inner  curve. 

136.    Other    Spirals.     The    possible    variety    of    spirals    is 
infinite.     In  general,  there  is  a  center  from  which  the  generating 


VII,  §  137] 


CURVED   LINES 


149 


point  moves  outward  in  increasing  convolutions.  The  law 
governing  the  generation  is  based  on  the  angle  through  which 
the  generator  moves  and  the  distance  it  attains  from  the  center. 

In  Fig.  146,  the  distances  0-1,  0-2,  etc.,  are  directly  propor- 
tional to  the  angles  9i,  02,  etc.  The  resulting  curve  is  the 
spiral  of  Archimedes.     (See  also  §  137.) 

Other  spirals  may  be  generated  by  making  the  distance  from 
the  center  proportional  to  the  square,  cube,  or  other  power 
of  the  angle.  Again 
the  distance  may 
be  inversely  propor- 
tional to  the  angle. 

137.  The  Helix. 
The  helix  is  a  space 
curve  (§§  108  and 
109)  and,  like  all 
other  curves  of  this 
class,  it  cannot  be 
drawn  on  a  flat  sheet 
of  paper.  It  may, 
however,  be  repre- 
sented by  its  projec- 
tions. 

A  helix  is   gener- 
ated by  a  point  mov- 
ing on  the  surface  of 
a  right  circular  cylinder,  in  such  a  manner  that  a  uniform  angular 
velocity  is  combined  with  a  uniform  velocity  paraHel  to  the  axis. 

In  Fig.  147,  ahcd  is  the  elevation  and  1,  5,  8,  12  is  the  plan 
of  a  right  circular  cylinder.  The  generating  point  starts  from 
1,  moving  to  the  left  around  the  cyhnder  and  rising  at  such  a 
rate  that  one  complete  revolution  is  made  in  the  height.  The 
construction  is  obvious. 

The  Hghter  line  shows  a  helix  that  makes  two  revolutions  in 


Fig.  146 


150 


DESCRIPTIVE   GEOMETRY 


[VII,  §  137 


the  same  height  and  moving  in  the  opposite  direction.     On  the 
front  of  the  cyHnder,  this  curve  ascends  toward  the  right,  and 

is    said    to    be    a   right-hand 
hehx,  with  a  pitch  of  1/2  p. 

It  will  be  seen  that  the 
instantaneous  direction  of  a 
helix,  with  respect  to  //,  is 
constant ;  i.e.  the  tangents  to 
a  helix  make  uniform  angles 
with  the  base  plane.  ^Vhen 
this  constant  angle  is  45°,  the 
F  projection  of  the  curve  will 
be  the  sinusoid  of  §  131. 
Examples  of  helixes  are  the 
thread  on  a  screw  or  bolt,  the 
hand  rail  on  a  spiral  stair, 
etc. 

Another  type  of  helix  can 
be  generated 
by  a  point 
moving  on  the 
surface  of  a 
cone,  follow- 
ing the  same 
laws  as  above. 
Such  a  curve  is  called  a  conical  helix.     Its  hori-  11 .  ■        h" 

zontal    projection   is    a   spiral   of   Archimedes. 
(See  §  136.) 

EXERCISE    SHEET    XLIX 

[Note.  Each  of  the  following  problems  is  intended 
to  be  solved  on  a  sheet  8"  X  10|",  with  margins  as 
shown  on  p.  xii.] 

The  door  shown  in  Fig.  148  is  hinged  at  h  and  h'. 
At  the  bottom  of  the  door  is  a  roller,  r,  at  each  side.  Fig.  148 


-IiLj: 


VII,  §  137] 


CURVED   LINES 


151 


When  the  dooi"  is  opened,  the  point  h'  moves  outward  and  upward  as 
indicated  by  the  arrow  h",  while  the  roller  moves  directly  upward,  in 
contact  with  the  wall,  as  shown  by  the  arrow  r' .  Determine  the  exact 
limits  of  the  space  not  interfered  with  by  the  opening  of  the  door. 
Scale  V'  =  I'-O". 

EXERCISE   SHEET  L 

In  Fig.  149  is  shown  a  casement  adjuster.  The  swivel  A  is  fast  to 
the  sill,  while  the  pivot  B  is  fast  to  the  sash.  The  rod  C  is  pivoted  at 
B  and  slides  in  A.  The  sash  is  hinged  at  D.  Lay  out  the  plan  as 
shown,  placing  it  on  left  hand  half  of  a  standard  sheet  divided  ver- 
tically. Place  hinge  3"  below  top  border.  Find :  1  (a)  The  path 
traced  by  the  end  of  the  rod  when  the  sash  is  opened.  1  (6)  The  posi- 
tion of  the  sash  when  its  swing  is  checked  by  the  rod.  2.  On  right 
half  of  sheet,  redraw  the  plan  and  let  the  adjuster  be  applied  to  the 
sash  and  sill  at  E  and  F,  instead  of  as  shown  in  Fig.  B.  Repeat  the 
above  solution.  Is  the  placing  described  in  1  better  or  worse  than 
that  in  2?     Scale  3"  =  I'-O". 


Elevation 
Fig.  149 


Llevatiorv 
Fig.  150 


EXERCISE   SHEET   LI 

Repeat  Problem  L  using  an  in-swinging  sash  adjuster  like  that 

shown  in  Fig.  150. 

EXERCISE   SHEET   LII 

(a)  Draw  both  of  the  projections  of  a  helix  generated  on  a  cylinder 
2"  in  diameter  and  3"  high,  and  draw  a  straight  line  tangent  to  the 
curve.  (6)  Draw  both  projections  of  the  curve  generated  on  a  sphere 
2"  in  diameter,  by  a  point  moving  with  uniform  angular  and  vertical 
velocities,  (c)  Draw  both  projections  of  the  curve  generated  in  the 
same  manner  on  a  cone,  2"  in  diameter  and  3"  high. 


152 


DESCRIPTIVE   GEOMETRY 


[VII,  §  137 


EXERCISE    SHEET   LUI 

(a)  Trace  the  epicycloid  formed  by  a  point  on  the  circumference 
of  a  circle  of  7/8"  radius  which  rolls  on  another  circle  of  If"  radius. 
(6)  Trace  the  hypocycloid,  using  the  same  base  circle,  but  letting  the 
rolling  circle  be  of  5/8"  radius. 

EXERCISE   SHEET  LIV 

Taking  the  same  circles  as  in  Sheet  LIII,  trace  the  path  of  a  point 
midway  between  the  center  and  circumference  of  the  moving  circle. 

EXERCISE    SHEET   LV 

The  derrick  in  Fig.  151  can  turn  on  the  vertical  mast  as  an  axis.  The 
distance  a  and  the  angle  6  may  be  made  to  vary.  Starting  with  ^  =  90'^ 
and  a  =  24'  — 0",  let  the  derrick  be  revolved  through  225°.  Meanwhile 
let  w  be  raised  till  a  =  2'— 0"  and  ^  =  30°.  Let  all  motions  be  uniform. 
Trace  the  path  of  w.     Scale  i-"  =  I'-O". 


Fig.  151 


•^777777Tr77i77777777777777TT7771-. 
Fig.  152 


EXERCISE   SHEET   LVI 


A  merry-go-round,  or  carrousel,  like  Fig.  152,  is  made  to  dip  twice 
in  a  revolution,  the  extreme  angle  of  inclination  of  the  ring  being  30°. 
Trace  the  motion  of  any  point  on  the  circumference,  assuming  vertical 
and  angular  velocities  to  be  uniform. 


Scale  i"  =  l'-0". 


EXERCISE   SHEET  LVII 

Place  the  sheet  vertically  with  UA  in  the  center.  Given  a  (33,  30, 
24),  and  6  (24,  14,  8).  Draw  the  projections  of  a  circle,  3"  in  diameter, 
whose  center  is  at  6,  and  whose  plane  is  perpendicular  to  ah. 

~"  EXERCISE  SHEET  LVIII 

In  the  center  of  the  sheet,  placed  horizontally,  draw  an  ellipse  whose 
major  axis  is  2"  and  whose  minor  axis  is  \" .  Place  the  major  axis 
vertically.     Draw  the  involute  (§  134).     Use  §  124  (3)  and  §  125. 


CHAPTER  VIII 
CURVED   SURFACES 

138.  Introduction.  The  use  of  curved  surfaces  in  the  form 
of  vaulting,  domes,  and  moldings  is  one  of  the  commonplaces 
of  architectural  practice.  The  sphere,  the  cylinder,  and  the 
cone,  are  the  most  usual  for  the  excellent  reason  that  they  are 
the  simplest.  It  is  not  so  evident,  however,  why  some  of  the 
other  forms,  notably  certain  warped  surfaces,  should  be  so 
rarely  used,  unless  it  be  that  designers  and  draftsmen,  through 
a  lack  of  knowledge  of  the  possible  forms  and  of  the  correct 
methods  for  generating  them,  are  timid  in  attempting  the 
unusual. 

A  good  knowledge  of  the  fundamental  ideas  regarding 
generation  of  surfaces  will  certainly  give  the  designer  a  broader 
means  of  expression,  and  an  abstract  study  of  surfaces  will 
introduce  him  to  some  rarely  encountered  and  inherently 
beautiful  forms  that  will  add  much  to  his  technical  vocabulary 
and  equipment. 

In  using  curved  surfaces,  some  discrimination  is  necessary 
in  selecting  materials  for  the  execution.  Materials  which 
are  plastic  while  being  worked,  such  as  terra  cotta,  or  plaster, 
may  be  used  for  nearly  any  describable  surface.  Bricks,  be- 
cause of  their  rectangular,  flat  faces  are  less  adaptable  to  small 
curved  surfaces ;  but  for  large  areas  the  numerous  joints  allow 
of  a  close  approximation  to  nearly  any  surface.  The  use  of 
flat  rectangular  tiles  to  approximate  a  large  variety  of  curved 
surfaces  is  a  notable  feature  of  the  Guasfavino  system  of  vaulting. 
Nearly  anything  can  be  done  in  stone  ;   but  the  cost  of  working 

is  high. 

153 


154  DESCRIPTIVE    GEOMETRY  [VIII,  §  139 

139.  Generation.  Just  as  a  line  is  generated  by  a  moving 
point,  a  surface  is  generated  by  a  moving  line.^  The  character 
of  any  surface  will  be  determined  by  the  following  items. 

(1)  The  character  of  the  moving  line  which  produces  it. 
This  line  is  called  the  generatrix  of  the  surface.  It  may  be 
straight,  or  it  may  be  curved  in  any  manner. 

(2)  The  particular  kind  of  motion  imparted  to  the  line. 
This  motion  may  be  a  motion  of  revolution,  or  a  motion  of 
transposition,  as  defined  below. 

Revolution  implies  an  axis,  i.e.  a  straight  line  about  which 
the  generatrix  revolves.  Each  point  of  the  generatrix  de- 
scribes a  circle,  the  plane  of  which  is  perpendicular  to  the  axis 
and  the  center  of  which  lies  in  the  axis.  The  essence  of  such 
a  motion  lies  in  the  fact  that  after  one  complete  revolution  the 
generatrix  comes  back  to  its  first  position.  Further  motion 
merely  retraces  the  surface  already  generated. 

A  surface  described  in  this  manner  is  called  a  surface  of 
revolution.  Such  a  surface  may  be  readily  executed  by  turning 
or  spinning  operations,  as  in  a  lathe  or  potters'  wheel. 

Transposition  is  any  motion  other  than  revolution.  Fre- 
quently it  is  a  motion  that  does  not  recur,  the  moving  line 
creating  a  surface  that  constantly  increases  as  long  as  the 
motion  is  continued.  In  this  type  of  motion  the  generatrix 
moves  so  as  to  constantly  touch  certain  guiding  lines,  called 
directrices;  or  so  as  to  maintain  a  fixed  relation  to  a  guiding 
surface,  called  a  director;    or  both. 

Any  one  position  of  the  generatrix  is  called  an  element  of  the 
surface. 

^  We  have  seen  how  the  motion  of  a  generatrix  (point  or  line)  in  a  direc 
tion  not  contained  within  itself  creates  a  quantity  (line  or  piano)  having 
one  dimension  more  than  its  genefatrix.  It  is  simple  to  see  how  a  two- 
dimensional  generatrix  gives  a  three-dimensional  solid.  The  student  who 
may  be  curious  to  know  what  would  happen  if  a  solid  could  be  moved  in  a 
direction  not  contained  in  itself  is  referred  to  "The  Fourth  Dimension 
Simply  Explained,"  by  Henry  P.  Manning;  and  to  an  interesting  applica- 
tion of  this  idea  in  "Projective  Ornament"  by  Claude  Bragdon. 


VIII,  §  140]  CURVED   SURFACES  155 

Surfaces  of  transposition  are  executed  by  ''  running  " ;  i.e. 
by  slipping  a  cutting  edge  along  guides  and  over  the  surface  to 
be  worked,  as  in  running  a  plaster  cornice ;  or  by  forcing  the 
material  past  a  revolving  knife,  as  in  a  molding  machine ;  or 
by  dies  or  rolls,  past  which  the  material  to  be  shaped  is 
drawn. 

As  an  example  of  the  use  of  the  preceding  terms,  a  sphere 
may  be  described  as  a  surface  of  revolution  whose  generatrix 
is  a  circle  that  rotates  about  an  axis  which  passes  through  its 
center  and  lies  in  its  own  plane.  Again  a  right  conoid  (see 
Fig.  175),  is  a  surface  of  transposition  generated  by  a  straight 
line  which  constantly  touches  both  a  given  semicircle  and  a 
given  straight  line  as  directrices,  while  remaining  parallel  to 
a  given  plane  director,  H. 

Any  surface  of  revolution  may  also  be  described  as  a  surface 
of  transposition.  Thus  a  right  circular  cylinder  may  be  gen- 
erated by  a  straight  line  revolving  about  an  axis  to  which  it 
is  parallel ;  or  by  a  circle  which  slides  along  a  straight  directrix. 
In  such  a  case  as  the  sphere,  the  generatrix  in  a  motion  of 
transposition  must  be  thought  of  as  variable  in  size. 

140.  Classification.  Since  the  character  of  any  curved 
surface  depends  upon  the  character  of  the  generatrix  and  the 
kind  of  motion  imparted  to  it  (§  139),  the  grouping  of  such 
surfaces  for  the  purpose  of  study  must  naturally  be  based  on 
these  two  considerations. 

We  shall  classify  surfaces  fbst  according  to  the  generatrix. 
Any  surface  that  can  be  generated  by  the  motion  of  a  straight 
line  is  called  a  ruled  surface.  All  others  are  called  double- 
curved  surfaces. 

Another  classification  is  by  means  of  the  motimi  of  the  gen- 
eratrix. Any  surface  formed  by  rotating  the  generatrix  as 
described  in  §  139,  is  called  a  surface  of  revolution.  All  others 
are  surfaces  of  transposition.  These  two  classifications  are 
independent  of  one  another.     Thus  there  may  be  ruled  surfaces 


156  DESCRIPTIVE   GEOIMETRY  [VIII,  §  140 

of  revolution  or  of  transposition ;  and  there  may  be  double 
curved  surfaces  either  of  revolution  or  of  transposition. 

The  characteristic  difference  between  ruled  and  double- 
curved  surfaces  lies  in  the  fact  that  while  a  ruled  surface  is 
composed  wholly  of  straight-line  elements,  and  therefore  a 
straight  line  can  be  drawn  through  any  point  of  the  surface, 
a  straight  line  cannot  be  drawn  through  an  arhitrarily  chosen 
point  on  a  double-curved  surface.  In  fnct,  the  more  usual 
double-curved  surfaces,  for  example  those  used  as  illustrations 
here,  can  contain  no  straight  line  whatever. 

141.  Tangents  and  Normals.  If  any  curved  surface,  ss  the 
sphere  in  Fig.  153,  is  cut  by  a  plane  P  which  passes  through 


Fig.  153 

a  given  point  a,  on  the  surface ;  the  tangent,  tt\  to  the  curve 
of  intersection  is  also  tangent  to  the  surface  at  the  given  point. 
Two  such  cutting  planes  passing  through  the  same  point  on  the 
surface  would  determine  two  intersecting  tangent  lines.  These 
lines  in  turn  would  determine  the  plane  which  is  tangent  to  the 
given  surface  at  the  given  point. 

In  the  case  of  a  ruled  surface,  a  tangent  plane  can  be  de- 
termined as  above  or  by  a  single  tangent  line  and  the  element 
of  the  surface  which  passes  through  the  given  point.     (See  §  161 .) 

A  line  is  normal  to  a  curved  surface  when  it  is  perpendicular 
to  a  tangent  plane  at  the  point  of  tangency.  A  plane  is  normal 
to  a  surface  if  it  contains  a  line  which  is  normal  to  the  surface. 

142.  Representation.  Curved  surfaces  lack  the  sharply 
limiting  points  and  lines  that  are  characteristic  of  plane  solids. 


VIII,  §  142] 


CURVED   SURFACES 


157 


Therefore  it  is  less  easy  to  visualize  their  projections.  However 
the  fundamental  idea  is  about  the  same.  Projecting  lines 
are  passed  tangent  to  the  surface,  forming  in  the  aggregate, 
an  envelope,  which  is  tangent  to  the  surface  as  shown  in  Fig. 
154.  The  line  of  tangency  between  a  surface  and  its  envelope 
is  called  the  line  of  sight.  In  general  the  envelope  will  take  the 
form  of  a  cylinder  (see  §  145  and  Fig.  157)  which  may  be  com- 
bined with  planes. 

Some  surfaces,  particularly  the  warped  surfaces,  can  be 
shown  best  by  drawing  the  projections  of  a  number  of  elements 
(Fig.  177). 

The  line  of  sight  on  any  curved  surface  is  the  boundary 
between  the  parts  of  the  surface  that  are  visible  and  those  that 


'The  envelope 


fVojecting  lines 


Fig.  154 


Projeciion  of 
the  sphere. 


are  invisible  in  projection.  It  is  well  to  note  that  some  parts 
of  a  surface  that  are  visible  on  the  H  projection  may  be  invisible 
on  the  V  projection,  and  vice  versa.  Also  there  are  other  parts 
that  may  be  visible  on  both  projections,  and  still  others  that 
may  be  invisible  on  both.  Care  should  be  taken,  in  drawing 
projections,  to  keep  the  visible  and  the  invisible  portions  clearly 
separated  in  the  mind. 


158 


DESCRIPTIVE   GEOMETRY  [VIII,  §  143 


143.  Ruled  Surfaces.  Ruled  surfaces  vary  widely  in  char- 
acter. Since  all  ruled  surfaces  have  the  same  generatrix  (§  140), 
this  variation  must  be  the  result  of  the  kind  of  motion  imparted 
to  the  generatrix.  Whether  in  any  given  case  the  motion  of 
a  generatrix  be  one  of  revolution  or  one  of  transposition,  its 
effect  on  the  character  of  the  surface  generated  can  be  best 


Upper  bosejN 


Upper  nappe 


Clement. 

Open  cone.  X/    /     i\| 

Lower  ^  nappe.    •        m 

0=Anqle  of  inclination 
.^' Lower  bo6e. 


Right  circular  cone. 


Right  circular  cylinder, 


Open ,  inclined  cylinder. 


Fig.  155 


studied  by  noting  the  relations  existing  between  adjacent  ele- 
ments. By  adjacent  elements  is  meant  two  elements  between 
which  there  is  no  appreciable  distance.^  All  of  the  possible 
relations  that  can  exist  between  adjacent  elements  can  be  sum- 
marized as  follows : 

(I)   Adjacent  Elements  in  the  Same  Plane. 

(1)  When  more  than  hvo  adjacent  elements  are  in  the  same 
plane,  the  surface  is  plane  or  plane-sided. 

1  See  footnote,  p.  120. 


VIII,  §  145]  CURVED   SURFACES  159 

(2)  When  only  two  adjacent  elements  lie  in  the  same  plane, 
the  surface  is  curved.     The  two  adjacent  elements  may  be : 

(a)  Parallel.     The  surface  is  a  cylinder  (Fig.  155). 

(6)  All  intersecting  at  the  same  point.  The  surface  is  a  cone 
(Fig.  155). 

(c)  Intersecting  two  by  two  (Fig.  171).  Since  the  elements 
are  adjacent,  the  points  of  intersection  will  be  adjacent ;  and 
since  no  three  elements  are  in  the  same  plane,  no  four  of  the 
points  of  intersection  will  be  in  the  same  plane.  Hence  these 
points  of  intersection  will  form  a  space  curve,  to  which  the  ele- 
ments are  tangents.  The  surface  thus  generated  is  called  a 
convolute. 

(II)  Adjacent  Elements  not  in  the  Same  Plane.  The 
surface  is  some  form  of  warped  surface. 

In  group  (I)  above,  any  two  adjacent  elements  determine 
a  plane.  Hence  the  surfaces  are  developable  for  the  same 
reasons  as  for  prisms  and  pyramids  (§  49).  Warped  surfaces, 
however,  are  not  developable  since  even  their  smallest  imaginable 
elements  are  not  planes  (Fig.  175). 

144.  Developable  Surfaces.  Developable  surfaces,  i.e.  all 
surfaces  included  under  heading  (I)  of  §  143,  play  an  important 
role  in  building  operations.  Wherever  sheet  metal  is  used, 
wherever  the  designs  used  are  set  in  mosaic  or  painted  on 
canvas ;  in  fact,  wherever  a  material,  originally  flat,  is  to  be 
applied  to  a  curved  surface,  it  is  important  that  the  surface 
in   question   be    developable. 

145.  The  Cone  and  the  Cylinder.  Generation.  A  cone 
is  generated  by  a  straight  line  which  always  passes  through 
a  fixed  point  (the  apex)  while  at  the  sa;me  time  it  moves  in 
contact  with  a  curved  directrix.  The  directrix  may  be 
any  kind  of  a  curve,  either  a  plane  or  a  space  curve  and 
it  may  be  either  open  or  closed.  If  the  generatrix  is 
unlimited,  the  cone  will  be  composed  of  two  equal  nappes 
meeting  at  the  apex  and  it  will  have  no  definite  base. 


160 


DESCRIPTIVE   GEOMETRY 


[VIII,  §  145 


Usually,  however,  only  one  nappe  is  drawn,  and  the  elements 
terminate  in  a  plane  base.  The  right  circular  cone  is  a  special 
case  (Fig.  155). 

A  cylinder  is  merely  a  special-case  cone,  the  apex  being  at  an 
infinite  distance  from  the  base.  Thus  all  the  elements  are 
parallel,  and  there  are  no  separate  nappes.  The  directrix 
may  be  of  any  form.  The  right  circular  cylinder  is  a  special  case 
(Fig.  155). 

Tlie  right  section,  sometimes  called  simply  the  section,  of 
a  cylinder  is  that  section  which  is  cut  from  the  surface  by  a 


e'  =  e 
Fig.  156  a 


e*<e 

Fig.  156  6 


plane  perpendicular  to  the  elements.  In  the  case  of  a  cone, 
the  right  section  is  perpendicular  to  the  axis.     (Compare  §  52.) 

The  cone  and  cylinder  being  so  closely  related,  any  operation 
that  can  be  performed  on  one  of  them  may  be  performed  on 
the  other  with  only  a  slight  change  in  the  method,  and  the 
results  of  such  operations  present  strong  analogies. 

The  Conic  Sections.  If  a  right  circular  cone  is  cut  by  a 
plane,  the  line  of  intersection  will   be  a  phine  (i:rve  called  a 


VIII,  §  145] 


CURVED   SURFACES 


161 


conic  section.  If  the  Inclination  of  the  cutting  plane  is  equal 
to  that  of  the  cone,  the  curve  will  be  a  parabola.  (See  Fig. 
156  a,  and  compare  with  §  112.)  In  the  figure,  the  plane  P 
being  parallel  to  the  element  ab,  it  is  readily  seen  that  the  curve 
must  be  an  open  one. 

If  the  angle  of  inclination  of  the  cutting  plane  is  less  than 
that  of  the  cone,  the  curve  is  an  ellipse.  (See  Fig.  1566  and 
§  112.)     When  the  inclination  of  the  cutting  plane  is  greater 


Fig.  157 


than  that  of  the  cone  the  curve  is  an  hyperbola.  (See  Fig.  156  c 
and  §  112.) 

From  the  figures  it  is  easy  to  see  that  the  ellipse  is  a  closed 
curve,  symmetrical  about  two  axes,  and  that  the  hyperbola  is 
an  open  curve  of  two  equal  branches. 

Projections.  The  envelope  (§  142)  of  a  cone  consists 
of  two  intersecting  planes  tangent  to  the  elements  of  sight, 
and  a  cylinder  which  is  formed  by  the  projecting  lines  pass- 
ing through  the  points  forming  the  base  of  the  cone.  (Fig. 
157.)  The  planes  and  the  cylinder  which  form  the  envelope 
are  tangent,  hence  the  projection  of  the  cone  will  consist  of  two 
intersecting  lines,  tangent  to  a  curve. 

In  the  case  of  a  cylinder  there  are  two  bases,  and  the  elements 
of  sight  are  parallel. 


162 


DESCRIPTI\:E  geometry  [VIII,  §  146 


146.    To  Assume  a  Cone  or  Cylinder. 

I.  Base  on  H  or  on  V.  The  base  may  be  assumed  as  any 
curve,  drawn  at  random  on  H  (or  V)  as  abed,  Fig.  158.  The 
apex,  0,  may  also  be  assumed  at  will.  The  V  projecting  planes 
of  the  lines  of  sight  (§  142),  will  have  their  H  traces  tangent 
to  the  curve  of  the  base,  as  a''a^  c^c'\     Thus  the  lines  of  sight 

with  respect  to  V,  oa  and  oc, 
are  located.  The  elements  of 
sight  with  respect  to  H,  pass 
through  0,  tangent  to  the  base, 
as  o^b^,  o^dJ". 

In  the  case  of  a  cylinder,  the 
elements  of  sight  would  be 
parallel. 

11.  Base  not  on  a  Plane 
OF  Projection.  Since  any 
two  curves  drawn  one  above 
the  other  on  H  and  V  are  the 
projections  of  some  curve  in 
space  (§  109),  a  base  may  be 
assumed  at  will.  Also  the  apex 
may  be  assumed  at  random.  Lines  drawn  from  the  apex 
tangent  to  the  base  will  complete  each  projection. 

III.  Cone  with  Given  Right  Section  and  Axis  Inclined 
AT  Given  Angles  to  H  and  to  V.  The  axis  can  be 
established  by  §  29,  and  a  plane  can  be  passed  perpendicular 
to  this  line  at  one  end  by  §  82.  This  establishes  the  base  plane, 
axis,  and  apex.  The  given  base  curve  can  then  be  drawn  on 
the  plane  by  counter-rabattement.  (See  §§  95  and  128.) 
Finally,  the  elements  of  sight  may  be  drawn  through  the  apex 
and  tangent  to  the  base. 

147.  To  Assume  an  Element  of  a  Cone.  In  Fig.  159  any  Hne 
drawn  through  d"  and  cutting  the  base,  as  o^d",  is  the  H  pro- 
jection of  an  element.     The  V  projection  of  the  point  c  lies  on 


VIII,  §  148] 


CURVED   SURFACES 


163 


the  V  projection  of  the  base  and  above  c\  as  c\  This  point 
together  with  o^  determines  the  V  projection  of  the  element  oc. 
It^'should  be  noted  that  o'c'  is  also  the  V  projection  of  an  ele- 
ment on  the  back  of  the 
cone,  the  plan  of  which  is 
shown  at  o^c'''. 

148.  To  Assume  a  Point 
on  a  Cone.  (1)  Firsl 
method.  In  assuming  a 
point  on  a  plane  (§  65),  it 
was  necessary  to  use  an 
auxiliary  line.  For  the 
same  reasons,  it  is  neces- 
sary to  use  an  auxiliary 
line  in  this  problem.  The 
natural  one  to  use  is  an 
element  of  the  cone.  The 
solution  then  consists  in 
assuming  an  element 
(§  147),  and  choosing  any 
point  on  it,  as  the  point 
b  on  the  element  oc. 
(Fig.  159.) 

(2)  Second  mcfJiod.  An- 
other method  consists  in 
assuming  any  point  within 
the    V   projection   of   the  p^^ '  ^-g 

cone  as  the  V  projection 

of  the  required  point,  as  a^'  in  Fig.  159.  Now  pass  an  auxil- 
iary H  plane  through  the  point  a^  It  will  cut  a 'circle  of 
diameter  m  n''  from  the  cone.  Next  find  the  H  projection 
of  this  same  circle  {m'^n^  in  Fig.  159).  The  H  projection  of 
the  point  a  lies  on  this  circle  and  directly  below  a%  i.e.  either 
at  the  point  a^  or  at  the  point  a'\  depending  on  whether  the 


164 


DESCRIPTIVE  GEO^IETRY  [\ail,  §  148 


assumed  point  is  chosen  on  the  visible  or  on  the  invisible  part 
of  the  cone. 

[Note,  If  the  cone  is  thought  of  as  a  surface  of  transposition 
(§  139 j,  the  circle  mn  is  an  element.  Thus  either  of  the  above  methods 
is  seen  to  depend  upon  determining  an  element  and  choosing  a  point 

on  that  element.] 

EXERCISE   SHEET   LIX 

1.  Given  a  (57,  20,  20),  h  (61,  6,  4),  and  c  (72,  8,  3).  Draw  the 
projections  of  an  open  cone,  similar  to  the  one  in  Fig.  155,  whose 
terminating  elements  are  ab  and  ac.  Draw  at  least  six  elements.  Let 
all  invisible  lines  be  dotted. 

2.  Draw  the  projections  of  a  right  circular  cylinder,  1"  in  diameter, 
the  axis  of  which  is  the  line  d  (32,  7,  6),  e  (39,  17,  10).     See  §  128. 

3.  The  point  /  (14,  0,  10)  is  the  center  of  the  base  of  a  right  circular 
cone.  If"  in  diameter,  the  elements  of  which  make  angles  of  60°  with 
H.  Locate  the  following  points  on  the  surface  of  the  cone :  g  and  h , 
4  from  H  and  13  from  V  ;  i  and  j,  4  from  H  and  7  from  T^ ;  k,  13  from 
P  and  6  from  H,  and  ???,  16  from  V  and  3  from  H. 

149.  To  Find  the  Line  of  Intersection  between  Any  Two 
Surfaces.     The  general  problem  of   intersection  is   extremely 

important.  It  arises 
constantly  in  connec- 
tion with  roofing  and 
vaulting.  A  large 
number  of  the  prob- 
lems in  stereotomy  and 
in  shades  and  shadows 
are  merely  intersection 
problems. 

It  has  been  pointed 
out  that  we  can  deal 

^        ^^  with  surfaces    in   pro- 

FiG.  160  .       .  ,     , 

jection  only  by  means 

of    lines    lying    in    the    surfaces.      This    idea  is   fundamental' 

The  use  of  auxiliary  planes  to  determine  the  required  lines  is 


VIII,  §  149] 


CURVED   SURFACES 


165 


familiar.  In  Fig.  160,  the  plane  A  cuts  from  the  cone  the 
circle  hcdmn.  From  the  sphere  the  plane  A  cuts  the  circle 
mnp.  The  points  iii  and  n,  where  these  circles  intersect,  are 
common  to  the  sphere  and  cone.  Hence  m  and  n  lie  on  the 
line  of  intersection  of  the 
surfaces. 

In  Fig.  161  the  same 
operation  is  shown  in 
projection.  The  points 
11  and  111  are  on  the  line 
of  intersection.  Other 
auxiliary  planes  passed 
parallel  to  A,  above  and 
below  it,  will  yield  other 
points.  When  a  suffi- 
cient number  of  points 
are  determined,  the  line 
of  intersection  is  drawn 
through  them.  The 
completed  curve  is  drawn 
in  Fig.  161. 

Slight  variations  of  the 
preceding  method  will  be 
sufficient  for  most  cases. 
It    is    sometimes    called 

the  slicing  mdhod.  The  necessary  variations  have  to  do  chiefly 
with  the  positions  of  the  cutting  planes.  It  should  be  recog- 
nized that  any  position  of  a  cutting  plane  will  yield  points  on 
the  required  line. 

It  is  obvious,  however,  that  the  solution  is  much  simplified 
if  the  lines  cut  by  the  auxiliary  planes  are  as  simple  as 
possible.  Hence  it  is  usual,  if  possible,  to  select  cutting 
planes  that  will  cut  circles  or  straight  lines  from  the  given 
surfaces. 


Fig.  161 


166 


DESCRIPTR^   GEOMETRY  [VIII,  §  150 


150.  To  Find  the  Line  of  Intersection  between  a  Plane  and 
a  Cylinder  (or  a  Cone). 

I.  Right  Circular  Cone  Resting  on  H.  In  Fig.  162, 
let  the  cone  and  the  plane  Z  be  given,  and  let  their  line  of 
intersection  be  required.     In  drawing  the  figure,  planes  parallel 


Fig.  162 


to  //  were  used  as  cutting  planes.  The  H  plane  cuts  from 
Z  the  trace  ZZ^,  and  from  the  cone  it  cuts  the  base  a^h^c^.  Thus 
a  and  h  are  determined  by  inspection  as  points  on  the  line  of 
intersection.  The  plane  A'  cuts  from  Z  the  line  vm.  From 
the  cone  the  plane  X  cuts  a  circle  whose  diameter  is  p^r^     This 


VIII,  §  150] 


CURVED  SURFACES 


167 


plane  yields  the  points  q  and  s,  on  the  line  of  intersection. 
Other  points  can  be  determined  by  using  other  planes.  The 
highest  point  (o)  on  the  required  curve  lies  on  that  line  of 
greatest  declivity  of  Z  which  passes  through  the  axis  of  the  cone. 
Repeat  this  construction  using  cutting  planes  that  pass 
through  the  apex  of  the  cone  and  are  perpendicular  to  H. 
Also  make  some  sketches 
showing  what  would 
happen  if  cutting  planes 
parallel  to  F  were  chosen. 

II.  Elliptical  Cyl- 
inder IN  ANY  Position. 
In  Fig.  163,  let  the 
cylinder  and  the  plane  Y 
be  given.  Horizontal 
cutting  planes  would  cut 
ellipses  from  the  cylinder. 
To  avoid  the  labor  of 
drawing  these  curves,  let 
cutting  planes  perpen- 
dicular to  H  and  parallel 
to  the  elements  of  the 
cjdinder  be  used,  as  W. 
This  plane  cuts  from  the 
cylinder  the  elements  ah 
and  cd  and  from  the 
plane  Y  it  cuts  the  line 
mn.     The  points  g  and  h  are  points  on  the  Hne  of  intersection. 

III.  A  Cone  in  any  Position.  This  case  is  much  like 
II,  above.  However,  it  would  probably  be  best  to  pass  the 
auxiliary  planes  perpendicular  to  ^  or  F  through  the  apex  of 
the  cone. 

[Note.     In  any  of  the  above  cases,  if  the  true  shape  of  the  line  of 
intersection  is  required,  it  may  be  determined  by  rabattement  (§93).] 


168  DESCRIPTIVE  GEOMETRY         [VIII,  §  150 


EXERCISE    SHEET   LX 

1.  The  point  a  (66,  0,  12)  is  the  center  of  a  circle  in  H,  whose  radius 
is  I".  This  circle  is  the  base  of  a  right  circular  cylinder.  The  cylinder 
is  cut  by  the  plane  Z  (67,  Z^  60°  I,  Z-  45°  r).  Find  the  Hne  of  inter- 
section between  the  plane  and  the  cylinder. 

2.  The  point  h  (28,  13,  0),  is  the  center  of  an  ellipse  in  V,  whose 
vertical  axis  is  2"  and  whose  horizontal  axis  is  1|".  This  ellipse  is  the 
base  of  a  cone  whose  apex  is  the  point  c  (39,  3,  23).  The  cone  is  cut 
by  the  plane  Y  (19,  Y^  30°  I,  Y^  75°  I).  Find  both  projections  of  the 
line  of  intersection  and  determine  its  true  shape  by  rabattement  into  T'. 

151.   To  Find  where  a  Line  Pierces  a  Cone  (or  a  Cylinder). 

I.  A  General  Method.  Pass  any  plane  through  the  given 
line  (§  66).  Find  the  line  of  intersection  between  this  plane 
and  the  cone  (§  150).  The  points  in  which  the  curve  thus 
found  intersects  the  given  line  are  the  required  piercing  points. 
The  theory  of  this  operation  is  identical  with  that  of  §  78. 

II.  Base  on  H  or  on  V.  If  the  cone  rests  on  i7  or  F,  a 
simpler  construction  than  that  just  given  is  possible.  Pass 
a  plane  through  the  line  and  through  the  apex  of  the  cone 
(§  69).  This  plane  w^ill  cut  two  elements  from  the  cone.  The 
points  in  which  these  elements  intersect  the  given  line  are  the 
required  piercing  points. 

[Note.  If  the  surface  in  question  is  a  cylinder  instead  of  a  cone, 
the  cutting  plane  is  passed  through  the  given  line  and  parallel  to  an 
element  of  the  cylinder.     Why  ?] 

EXERCISE   SHEET  LXI 

1.  The  point  a  (69,  0,  11)  is  the  center  of  a  circle,  If"  in  diameter 
and  which  lies  in  H.  The  hne  a,  b  (50,  19,  22)  is  the  axis  of  a  cylinder 
whose  base  is  the  given  circle.  The  line  c  (72,  15,  24),  d  (51,  6,  3) 
pierces  the  cylinder.     Find  the  piercing  points. 

2.  The  point  e  (14,  14,  22)  is  the  center  of  a  circle.  If "  in  diameter^ 
which  lies  in  a  plane  parallel  to  V.  This  circle  is  the  base  of  a  cone 
whose  apex  is  at/  (32,  3,  3).  Find  the  points  where  the  line  g  (9,  5> 
4),  h  (28,  12,  19),  pierces  the  cone. 


VIII,  §  152] 


CURVED  SURFACES 


169 


152.  To  Find  the  Line  of  Intersection  between  Cones  and 
Cylinders.  It  is  impossible  to  state  any  one  best  method. 
The  choice  of  a  trpe  of  cutting  plane  is  all  important.     In 

b"  d^ 


Fig.  164 

general,  the  best  plane  to  choose  is  that  which  cuts  the  simplest 
lines  from  the  given  surfaces. 

In  Fig.  164,  if  the  bases  were  circles,  planes  parallel  to  H 
would  do  very  well ;  but  with  elliptical  bases,  probably  it  would 
be  easier  to  use  planes  parallel  to  the  axes  of  both  cyhnders 
(§  72).  In  the  figure  a  plane  R  is  used.  The  V  trace  of  R  is 
not  drawn,  since  the  H  trace  is  sufficient  to  establish  the  ele- 
ments ah  and  c^  on  one  cylinder,  and  cf  and  gh  on  the  other. 
These  elements  give  four  points,  1-4,  on  the  line  of  intersection. 


170 


DESCRIPTIVE   GEOMETRY  [VIII,  §  152 


In  the  case  shown  in  Fig.  165,  planes  passed  through  both 
apexes  will  cut  elements  from  both  cones.  An  auxiliary  line 
passed  through  the  apexes,  as  vin  is  used  to  determine  the 
planes  (§  66).  In  the  figure,  one  plane,  W,  was  used  as  an 
example.     The  V  trace  is  omitted,  as  in  Fig.  164. 


Fig.  165 


In  Fig.  166,  the  position  of  the  objects  is  such  that  an  auxiliary 
projection  on  a  V  plane,  parallel  to  the  base  of  the  cylinder, 
is  found  most  convenient.  Once  this  projection  is  found, 
planes  passed  through  the  apex  and  parallel  to  elements  of  the 
cylinder,  as  X,  will  cut  straight  lines  from  each  surface. 


VIII,  §  152] 


CURVED  SURFACES 


171 


Fig.  166 


172 


DESCRIPTIVE   GEOMETRY         [VIII,  §  152 


When  both  surfaces  are  surfaces  of  revolution  and  their 
axes  intersect,  as  in  Fig.  167,  concentric  spheres  with  their 
centers  at  the  intersection  of  the  axes  can  be  used.  This  case 
is  more  fully  treated  in  §  177,  II. 


Fig.  167 


It  will  be  seen  from  the  above  examples  that  much  ingenuity 
can  be  exercised  in  selecting  an  auxiliary  plane  or  other  surface 
that  suits  the  given  conditions.  A  wise  selection  is  possible 
only  when  the  surfaces  to  be  dealt  with  are  thoroughly  under- 
stood and  a  number  of  cases  have  been  worked  out  carefully. 


VIII,  §  152]  CURVED   SURFACES  173 

[Note.  Rapidity  and  accuracy  in  intersection  problems  can  be 
attained  only  if  a  careful  system  of  notation  is  used.  It  is  well  also 
to  think  out  carefully  not  only  the  kind  of  plane  to  be  used,  but  also 
the  most  advantageous  spacing,  whether  regular  or  irregular.  Often 
some  points  can  be  determined  by  inspection;  where  possible,  this 
should  be  done.  Visualize  the  required  line  as  completely  as  possible. 
If  one  surface  is  symmetrical,  start  from  the  center  and  work  both 
ways.  After  drawing  an  auxiUary  plane,  determine  the  required  points 
both  on  plan  and  elevation  before  another  plane  is  drawn. 

In  each  of  the  preceding  cases,  the  Hne  of  intersection  is  tangent 
to  the  elements  of  sight.  This  is  true  for  any  problem  in  intersections 
where  the  entire  surface  is  cut.] 

EXERCISE   SHEET   LXII 

1.  The  line  a  (70,  0,  13),  h  (54,  16,  13)  is  the  axis  of  a  U"  right 
circular  cylinder.  The  line  c  (56,  0,  14),  d  (73,  13,  14)  is  the  axis  of  a 
1"  right  circular  cylinder.     Find  the  line  of  intersection  of  the  surfaces. 

2.  The  Hne  e  (30,  0,  8),  /  (6,  20,  13)  is  the  axis  of  an  elhptical  cone, 
the  base  of  which  is  a  2"  circle  in  H.  The  hne  g  (12,  0,  5),  h  (34, 18,  18) 
is  the  axis  of  an  elliptical  cylinder,  the  base  of  which  is  a  Ij"  circle  in 
H.     Find  the  line  of  intersection  of  the  surfaces. 


EXERCISE   SHEET   LXIII 

1.  The  point  a  (62.  0,  12)  is  the  center  of  a  2\"  circle  in  H.  This 
circle  is  the  center  of  the  base  of  a  cone  whose  apex  is  at  h  (72,  23,  3). 
A  right  circular  cone,  1|"  in  diameter  and  If"  high,  has  the  center  of 
its  base  at  c  (59,  0,  13).     Find  the  line  of  intersection  of  the  surfaces. 

2.  The  line  d  (13,  0,  21),  e  (28,  22,  6)  is  the  axis  of  a  cyhnder  whose 
base  is  a  cu-cle  If"  m  diameter  in  H.  The  line/  (30,  0,  19),  g  (16,  22, 
5)  is  the  axis  of  a  cyhnder  whose  base  is  a  circle  1\"  in  diameter  in  H. 
Find  the  line  of  intersection  of  the  surfaces. 


EXERCISE   SHEET   LXIV 

Place  sheet  vertically,  HA  in  the  center.  The  point  a  (28,  0,  20) 
is  the  center  of  the  base  of  a  right  cu-cular  cone,  4"  in  diameter  and  4^" 
high.  The  line  h  (52,  16,  20),  c  (4,  16,  20)  is  the  axis  of  a  right  circular 
cylinder  3"  in  diameter.     Find  the  line  of  intersection  of  the  surfaces. 


174 


DESCRIPTIVE  GEOMETRY 


[VIII,  §153 


153.  To  Pass  a  Plane  Tangent  to  a  Cone  through  a  Point  on 
the  Surface.     (See  §  141.) 

I.  Base  of  Cone  on  H  or  on  V.  In  Fig.  168,  let  the  point 
a  be  determined  on  the  surface  of  the  cone  as  in  §  148.  Let  it 
be  required  to  pass  a  plane  through  a  tangent  to  the  cone.  Any 
tangent  plane  which  contains  the  point  a  will  also  contain  the 
element  he,  which  passes  through  a.  The  H  trace  of  the  tangent 
plane  is  tangent  to  the  cone  base  at  h^,  as  ZZ^.  The  V  trace 
passes  through  the  V  trace,  d^\  of  the  element  of  contact. 

II.  Base  of  Cone  not 
z>^;"  ON  H  NOR  ON  F.  Find 
the  element  of  contact  as 
before.  Pass  an  auxiliary 
H  (or  V)  plane  through 
the  given  point  and  find 
its  line  of  intersection 
with  the  cone.  Draw  a 
line  in  this  plane  which  is 
tangent  to  the  line  of  inter- 
section, and  which  passes 
through  the  given  point. 
This  tangent  and  the  ele- 
ment of  contact  will  deter- 
mine the  required  plane. 
Sometimes  it  is  easier  to  extend  the  cone  to  H  (or  to  V)  and 
to  proceed  as  in  I  above.  The  idea  is  the  same  in  either  case. 
III.  Cylinders.  The  same  construction  will  suffice  for 
cylinders  if  it  is  kept  in  mind  that  a  cylinder  may  be  considered 
as  a  cone  with  its  apex  at  an  infinite  distance. 

154.  To  Pass  a  Plane  Tangent  to  a  Cone  (or  a  Cylinder) 
through  a  Point  Outside.  Every  tangent  plane  must  pass 
through  the  apex  of  the  cone.  Hence  a  line  passed  through  the 
apex  and  the  given  point  will  be  in  the  required  plane.  The 
traces  of  the  required  plane  pass  through  the  traces  of  this  line. 


Fig.  168 


VIII,  §  155] 


CURVED   SURFACES 


175 


//  the  cone  rests  on  H,  the  H  trace  of  the  required  plane  will 
be  tangent  to  the  base  of  the  cone.  Two  tangent  planes  may 
be  passed  through  any  point  outside  the  surface. 

//  the  hose  of  the  cone  does  not  rest  on  H,  an  auxiliary  plane 
parallel  to  //,  and  cutting  the  cone  can  le  used,  as  in  §  153,  II. 

//  the  given  surface  is  a  cylinder  {i.e.  a  cone  with  its  apex  at 
an  infinite  distance),  the  only  change  in  the  preceding  method 
consists  in  drawing  the  guide 
line  through  the  given  point  / 
parallel  to  an  element. 

155.   To    Pass    a    Plane 
Tangent  to  a  Cylinder  (or 
a  Cone)  and  Parallel  to  a 
Given  Line.     Such  a  plane 
must  contain  an 
element   of    the 
cylinder;  hence 
it  will  be  paral- 
lel to  all  of  the 
elements.        In 
Fig.  169,  let  ab  be  the  given 
line.     Through   any   point 
on   ab,  as   c,  draw   a    line 
parallel  to   an  element  of 
the    cylinder,  as  cd.     The 

lines  ab  and  cd  determine  a  plane,  X.  Any  plane  parallel  to 
X,  the  H  trace  of  which  is  tangent  to  the  base  of  the  cylinder 
as   Y  or  Z,  will  contain  an  element  and  will  be  parallel  to  ab. 

If  the  given  surface  is  a  cone,  a  guiding  line  passed  through 
the  apex  and  parallel  to  the  given  line  may  be  used.  The 
traces  of  this  line  give  one  point  on  each  of  the  traces  of  the 
required  plane.  The  H  trace  must  also  be  tangent  to  the  base 
of  the  cone.  If  the  inclination  of  the  given  line  toward  H  is 
less  than  that  of  the  cone,  there  are  two  possible  tangent  planes. 


Fig.  169 


176  DESCRIPTR^  GEOMETRY  [VIII,  §  155 

If  the  inclination  of  the  given  Hne  is  equal  to  that  of  the  cone, 
there  is  but  one  tangent  plane.  If  the  inclination  of  the  line 
is  greater  than  that  of  the  cone,  the  construction  is  impossible. 

156.  To  Pass  a  Line  Tangent  to  a  Cone  or  a  Cylinder.  Fol- 
lowing the  general  methods  heretofore  employed,  two  different 
methods  are  apparent. 

(a)  Pass  a  plane  cutting  the  given  surface  in  a  curve.  Draw 
a  line  i7i  that  p^.ane  tangent  to  the  curve  (§  141). 

Q))  Pass  a  plane  tangent  to  the  surface,  and  draw  in  that 
plane  the  required  tangent  line. 

EXERCISE   SHEET  LXV 

1.  Given  a  45°  right  circular  cone,  If"  in  diameter,  resting  on  H 
and  with  the  center  of  its  base  at  the  point  a  (67,  0,  10).  Determine 
two  points,  h  and  c,  on  the  surface  of  the  cone,  64  from  P  and  3  from 
H.  Draw  the  traces  of  the  planes  X  and  F,  which  are  tangent  to  the 
cone  at  the  points  h  and  c. 

2.  The  point  d  (40,  10,  11)  is  the  center  of  a  \\"  circle  parallel  to  V. 
This  circle  is  the  base  of  an  oblique  cone  whose  apex  is  at  the  point  g 
(33,  4,  3).  Determine  the  points  e  and  /,  which  He  on  the  cone,  14 
from  H  and  11  from  V.  Through  e  and/,  draw  the  planes  V  and  W, 
tangent  to  the  cone. 

3.  The  line  h  (4,  0,  5),i  (11,  13,  16)  is  the  axis  of  an  oblique  cylinder 
whose  base  is  a  1"  circle  in  H.  Through  the  point  k  (13,  5,  10),  draw 
the  planes  T  and  U,  tangent  to  the  cylinder. 

EXERCISE   SHEET  LXVI 

1.  Given  a  60°  right  circular  cone,  \\"  in  diameter,  resting  on  H 
and  with  the  center  of  its  base  at  a  (70,  0,  11).  Given  also  the  lines 
h  (59,  0,  7),  c  (65,  13,  11),  and  d  (74,  23,  18),  e  (75^,  15,  23).  Draw  as 
many  planes  as  possible  which  are  tangent  to  the  cone  and  parallel  to 
either  6c  or  de. 

2.  The  line/  (45,  0,  11),  g  (39, 12,  22)  is  the  axis  of  an  oblique  cyhnder 
whose  base  is  a  l\"  circle  in  H.  Through  the  point  h  (33,  8,  4)  draw 
two  planes  tangent  to  the  cylinder. 

3.  The  line/  (16,  0,  17),  k  (8,  16,  8)  is  the  axis  of  an  obHque  cyhnder 
whose  base  is  a  Ij"  circle  in  H.    Through  the  point  I  (14,  17,  1)  draw 


YIII,  §  157] 


CURVED  SURFACES 


177 


three  lines,  Im,  In,  and  lo,  which  are  tangent  to  che  cyHnder  at  points 
^",  1",  and  1|",  respectively,  above  H.     (See  §  156  b.) 

157.    To  Develop  a  Cone. 

I.  Right  Circular  Cone.  If  a  cone  is  placed  on  a  plane 
with  one  element  of  the  cone  in  contact  with  the  plane,  and  the 
cone  is  then  rolled  out  on  the  plane,  the  original  element  mean- 
while remaining  fixed,  the  resulting  figure  is  the  development 
of  the  cone  (§  49). 

Since  all  the  elements  of  the  cone  are  of  equal  length,  and  the 
apex  does  not  move,  the  development  will  be  a  sector  of  a 
circle.  The  radius 
of  this  sector  will 
be  the  slant  height 
of  the  cone.  The 
actual  length  of  the 
arc  bounding  the 
sector  is  equal  to 
the  circumference 
of  the  cone  base. 
In  making  the 
drawing  the  length 
of  this  arc  may  be 
stepped  off  with 
the  dividers,  as  in  rectifying  a  cu^ve  (§  111),  or  the  angle  of 
the  sector  may  be  computed  by  proportion. 

Let  the  student  prove  that  the  angle  of  the  sector  given  by 
developing  a  right  circular  cone  is  equal  to  360°  cos  a,  where 
a  is  the  angle  of  inclination  of  an  element  of  the  cone  to  the 
base. 

II.  Oblique  Cone.  Between  any  two  elements,  drawn  near 
together  (as  oa,  oh,  Fig.  170),  the  surface  of  the  cone  ap- 
proximates a  triangle.  By  finding  the  true  lengths  of  its 
sides,  any  such  triangle  can  be  drawn  as  part  of  a  develop- 
ment,  as    at   oab.      This    method    is    a    more    or    less    close 


178  DESCRIPTIVE  GEOMETRY         [VIII,  §  157 

approximation,  depending   on   the   care   and   the   number   of 
elements  used. 

III.  Lines  on  the  Surface.  Any  line  on  the  surface  of 
a  cone  can  be  shown  on  the  development  by  finding  the  true 
lengths  of  the  intersected  elements  and  transferring  these  to 
the  development. 

If  it  is  required  to  trace  the  shortest  path  between  two  points 
on  a  surface,  locate  the  points  on  the  development,  draw  a 
straight  line  between  them  cutting  various  elements,  locate 
these  points  of  intersection  on  the  surface,  and  draw  a  curve 
through  them. 

IV.  Cylinders.  The  principles  and  methods  are  the  same 
as  for  the  cone. 

V.  To  Develop  a  Space  Curve.  (See  also  §  111,  II.)  Con- 
sider the  given  curve  as  the  generatrix  of  a  cone  or  cylinder. 
Draw  the  projections  of  the  surface.  This  surface  may  then 
be  developed  and  the  given  curve  traced  on  the  developments,  as 
in  III,  above.  If  required,  this  curve  may  then  be  rectified 
into  a  straight  line  as  in  §  111. 

EXERCISE  SHEET  LXVH 

[Note.  In  this  sheet,  and  in  several  hereafter,  no  special  location 
is  given  for  the  objects  to  be  drawTi.  In  such  cases  the  student  should 
carefully  plan  the  layout  from  the  start,  to  make  sure  that  all  the  re- 
quired work  can  be  done  without  overlapping,  and  that  the  finished 
sheet  will  be  attractively  arranged.  1 

1.  Draw  the  projections  of  a  60°  right  circular  cone,  whose  base  is 
3"  in  diameter.  2.  Develop  the  cone.  3.  Midway  on  the  center  ele- 
ment of  the  development,  draw  a  |"  square  and  trace  the  involute 
(§134)  of  this  square  on  the  development.  4.  Transfer  the  involute 
just  drawn  from  the  development  to  the  projections  of  the  cone. 

158.    The  Developable  Helicoid. 

I.  Type  Characteristics.  As  a  single  representative  of 
the  class  of  curves  defined  in  §  143, 1  (2,  c)  the  developable  helicoid 
will  be  chosen  for  study. 


VIII,  §  158]  CURVED  SURFACES  179 

Such  a  surface  is  generated  by  a  straight  Hne  generatrix  that 
moves  so  that  consecutive  positions  intersect  at  adjacent 
points,  no  four  of  which  taken  in  succession  He  in  the  same 
plane.^  In  Fig.  171,  let  ae  be  a  series  of  such  points.  Then 
it  is  evident  that  bg  and  ch  are  Hues  in  the  same  plane,  as  are 
also  ch  and  di. 

But  if  the  points  a  and  d  are  not  in  the  same  plane,  it  is 
evident  that  the  lines  bg  and  di  are  not  in  the  same  plane. 
Nevertheless  the  surface  afje  is  developable,  since  it  is  com- 
posed of  plane  elements,  as  in  the  case  of  the  pyramid. 

If  the  points  a  and  b,  b  and  c,  etc.,  are  made  to  approach  one 
another,  keeping  their  relative  positions,  the  broken  line  ae 
approaches  a  space  curve  to  which  the  lines  bg,  ch,  etc.,  are 

e 

d 


Fig.  171 

tangent ;  and  the  series  of  plane  surfaces  approaches  a  develop- 
able ruled  surface.  This  evidently  means  that  such  a  surface 
may  be  generated  from  any  space  curve. 

Such  a  surface  as  described  above,  being  a  developable  ruled 
surface,  is  most  closely  related  to  the  cone  and  cylinder  in  its 
construction,  but  it  bears  a  visually  closer  resemblance  to  some 
of  the  warped  surfaces  (§  161)  in  that  it  has  no  simple  visual 
outlines.  In  projection  the  shape  may  be  revealed  by  Hmiting 
the  surface  between  fixed  planes  which  the  surface  intersects, 
and  by  drawing  regularly  spaced  elements  on  plan  and  elevation. 

1  See  footnote,  p.  120. 


180 


DESCRIPTIVE   GEOMETRY         {VIU,  §  158 


11.  Special  Characteristics.  The  developable  helicoid  is 
generated  by  the  tangents  to  a  helix  (§  137).  If  the  tangent 
is  considered  as  extending  in  both  directions  from  the  curve, 
the  surface  which  is  generated  is  one  of  two  nappes.  If  the 
tangents  extend  in  but  one  direction,  the  surface  is  of  a  single 
nappe.  It  is  the  surface  that  results  from  unwrapping  a  string 
wound  about  a  cylinder  and  following  a  helix  (Fig.  172). 

Some  of  the  more  interesting  characteristics  of  this  surface 
can  be  seen  in  Fig.  172  (2).  Here  the  surface  is  limited  by  a 
pair  of  horizontal  planes,  the  spacing  of  which  is  equal  to  the 

(2) 

b 

h 


One  noppe. 


Two  nappes. 


Fio.  172 


pitch  of  the  helical  directrix.  The  tangent  at  a  is  continued 
to  the  limiting  planes,  giving  the  line  be.  From  the  manner 
in  which  a  helix  is  generated,  it  is  known  to  be  a  curve  of  constant 
inclination.  Hence  the  inclination  of  the  tangent,  i.e.  the  angle 
abd,  is  the  same  for  any  point  on  the  helix.  Therefore  the  lines 
be,  efy  gh,  etc.,  are  equal,  and  each  of  them  is  equal  to  the  length 
of  the  helix.  Moreover  the  H  projection  of  any  one  of  these 
lines  (fe)  will  be  tangent  to  the  //  projection  of  the  cylinder, 
and  the  length  of  the  projection  (gf)  will  be  equal  to  the  circum- 
ference of  the  circle  (glcx).  The  curve  gcf  (or  cbh),  as  well  as 
any  section  cut  from  the  surface  by  any  horizontal  plane,  is  an 
involute  of  the  circle  (§  134)  which  the  plane  cuts  from  the 
cylinder.     These  facts  may  be  visualized  without  formal  proof. 


VIII.  §  159] 


CURVED  SURFACES 


181 


159.  To  Draw  the  Projections  of  a  Developable  Helicoid  of 
Two  Nappes.  Let  the  directing  cylinder  and  one  turn  of  the 
helical  directrix  ahcde  be  given,  as  shown  in  Fig.  173.  Draw 
the  involute  (§  134)  of  the  lower  base,  I''— 4\  and  that  of  the 
upper  base,  w^ — z^.     Next  draw  tangents  equally  spaced  around 


Fig.  173 

the  base  circle,  and  extending  to  the  involutes,  as  X^h^y^,  2''cV, 
etc.  These  lines  are  the  H  projections  of  elements  of  the 
required  surface  (corresponding  to  fg  in  Fig.  172  (2)).  The 
V  projections  of  1,  2,  3,  4  are  in  HA,  and  those  of  ic,  x,  y,  z 
are  in  the  plane  of  the  upper  base,  WA\  The  elements  of 
both  nappes  can  now  be  drawn  on  the  V  projection,  as  a^'2% 


182 


DESCRIPTIVE   GEOIVIETRY  [VIII,  §  159 


If  accurately  determined,  the  V  projections  of  the  elements 
should  all  be  tangent  to  the  given  V  projection  of  the  helical 
directrix.  Or  if  the  V  projection  of  the  helix  has  not  been 
drawn,  these  tangents  will  determine  it. 

160.  To  Develop  the  Developable  Helicoid  of  One  Nappe. 
Consider  the  surface  shown  in  Fig.  174  a  as  an  approximation 


(6) 


Fig.  174 


to  a  developable  helicoid.  The  lines  ab,  be,  etc.,  are  equal, 
and  the  angles  abc,  bed,  etc.,  are  equal.  But  the  lines  aj,  jk, 
etc.,  are  not  equal,  nor  are  the  angles  ajk,  jkl,  etc.  If  the  surface 
is  developed,  the  line  a-i  becomes  a  broken  line,  the  component 
parts  of  which  are  equal  in  length  and  in  change  of  direction. 
The  line  a-q  becomes  an  irregular  broken  Hne  whose  elements 


VIII,  §  161]  CURVED   SURFACES  183 

increase  in  length.  The  angles  between  them  increase  in 
magnitude  as  the  line  progresses.  If,  by  decreasing  the  dis- 
tances ab,  be,  etc.,  the  surface  approaches  a  developable  helicoid, 
the  development  (Fig.  174  c/)  will  be  bounded  by  a  circular 
arc  ai-ii  and  the  curve  ai-gi. 

It  can  be  shown  by  trigonometry  that  the  radius,  a*,  of  the 
inner  curve  (Fig.  174  c/)  is  equal  to  R/cos-O,  where  R  is  the 
radius  of  the  cylinder  and  6  the  angle  of  inclination  of  the  helix. 
A  graphical  construction  for  determining  this  ratio  is  shown 
in  Fig.  174  e.  It  is  also  evident  that  the  length  of  this  arc 
ajiii  must  be  equal  to  the  length  of  the  helix  a^fi^  in  Fig.  174  6. 
The  required  length  may  be  found  by  developing  the  original 
helix  as  shown  in  Fig.  174  c,  and  measuring  the  distance  thus 
found  along  the  circumference  ajiii  Fig.  174  c/.  From  the 
manner  in  which  the  surface  is  generated,  it  is  evident  that 
the  length  of  any  element  f7i^  (Fig.  174  b)  is  equal  to  the  length 
of  the  curve  fd^a^.  Hence,  in  the  development,  the  length  of 
any  tangent  to  the  circular  arc/i7?i  must  be  equal  to  the  length 
of  the  arc  fidiOi.  From  this  consideration,  the  outer  curve  is 
seen  to  be  the  involute  (§  134)  of  the  inner  curve.  If  accurately 
constructed,  the  length  of  the  curve  aiUiQi  will  be  found  to 
equal  that  of  a^n}q^  in  Fig.  174  b. 

EXERCISE   SHEET  LXVIII 

1.  The  point  a  (67,  0,  20)  is  the  center  of  the  base  of  a  right  circular 
cylinder  1"  in  diameter  and  2"  high.  Draw  the  projections  of  the 
helix  and  the  developable  helicoid,  making  one  turn  about  the  cyUnder. 
With  the  point  b  (14,  0,  10)  as  the  center  of  the  inner  curve,  draw  the 
development  of  the  surface  just  drawn. 

161.  Warped  Surfaces.  I.  Generation  and  Character- 
istics. As  noted  in  §  143,  warped  surfaces  are  generated 
by  the  motion  of  a  straight  line  generatrix,  the  one  neces- 
sary condition  being  that  adjacent  positions  of  the  generatrix 
shall  not  lie  in  the  same  plane. 


184 


DESCRIPTIVE  GEOAIETRY 


[VIII,  §  161 


A  glance  at  Fig.  175,  which  represents  a  typical  warped  sur- 
face, will  show  that  adjacent  elements  (as  ab  and  cd)  are  not 
parallel ;  neither  can  they  intersect,  since  they  are  both  parallel 
to  the  same  plane. ^  That  portion  of  a  warped  surface  which 
lies  between  any  two  elements  can  not  be  approximated  by 
a  single  plane  figure,  as  is  the  case  with  cones  and  cylinders 
(§  157,  II).  The  details,  1  and  2,  in  the  figure,  show  how 
such  a  portion  of  a  warped  surface  can  be  approximated  by  two 
triangles  that  are  not  in  the  same  plane.     Thus  the  smallest 

imaginable  part  of  a  warped  sur- 
face is  not  a  plane,  and  it  follows 
that  the  whole  surface  is  non- 
developable.  An  approximate 
development  of  a  warped  surface 
may  be  made  by  triangulation, 
as  indicated  in  the  figure  and  as 
explained  in  §  157,  II. 

11,  Generation  by  Revolu- 
tion (§139).  A  ruled  surface 
of  revolution  is  formed  by  re- 
volving a  straight  line  about  an 
axis.  The  generatrix  must  bear 
one  of  three  possible  relations  to  the  axis.  It  may  be  (a)  paral- 
lel to,  (b)  intersecting,  or  (c)  not  in  the  same  plane  as  the  axis. 
In  cases  (a)  and  (6),  the  resulting  surfaces  are  the  cylinder  and 
cone.  Case  (c)  gives  an  hyperboloid  of  revolution,  which  may  be 
defined  as  a  surface  generated  by  a  straight  line,  revolving  about 
another  line  not  in  its  own  plane.  Since  these  three  are  the  only 
possible  cases  of  revolution  of  a  straight  line,  and  since  two  of 
them  give  developable  surfaces,  it  is  evident  that  the  hyperboloid 
of  revolution  is  the  only  possible  warped  surface  of  revolution. 
III.  Generation  by  Transposition  (§  139).  The  motion 
of  the  straight  line  generatrix  is  controlled  by  lines  or  planes. 

1  See  footnote,  p.  120. 


VIII,  §  162]  CURVED   SURFACES  185 

The  more  common  arrangements  of  the  directrices  are  as 
follows :  (1)  three  lines,  either  straight  or  curved,  or  straight 
and  curved  Hues  in  combination ;  or  (2)  two  lines  either  straight 
or  curved,  or  straight  and  curved  lines  in  combination,  and 
a  plane  director  to  which  the  generatrix  maintains  a  fixed 
relation.  The  number  of  ways  in  which  these  factors  can  be 
combined  is  very  great.  Hence  the  possible  number  of  warped 
surfaces  of  transposition  is  practically  unlimited.  A  few 
typical  cases  have  been  selected  for  study.  Once  these  are 
mastered,  no  difficulty  will  be  found  in  handling  other  cases. 

IV.  Time  as  a  Factor  ix  Generation.  Sometimes  it  is 
convenient  to  think  of  the  generation  of  a  warped  surface  as 
being  controlled  by  a  time  factor,  or  velocity  factor,  which 
replaces  one  of  the  geometric  factors  (line  or  plane)  mentioned 
above.  Thus  any  two  lines,  together  with  a  definite  rate  of 
motion  of  the  generatrix  along  each,  constitute  a  satisfactory 
method  of  defining  a  surface  which  is  often  more  convenient. 
For  example,  in  Fig.  177,  the  surface  is  generated  by  the  line 
ac  sliding  along  ah  and  cd,  in  equal  spaces  of  time  and  at  uniform 
velocities.     (See  also  §  201.) 

V.  Tangency.  a  line  tangent  to  a  warped  surface  is  de- 
termined as  explained  in  §  141.  A  plane  is  said  to  be  tangent 
to  a  warped  surface  when  it  contains  an  element  of  the  surface 
and  a  tangent  line.  By  reference  to  Figs.  176  and  177  6,  it  will 
be  seen  that  any  such  plane  will  intersect  the  surface  to  which 
it  is  said  to  be  tangent.  But  in  such  a  case  as  Fig.  175,  a  plane 
so  determined  will  be  tangent  in  the  usual  sense. 

162.  Representation.  In  purely  theoretical  discussions, 
warped  surfaces,  as  well  as  planes,  cones,  etc.,  are  supposed 
to  be  of  indefinite  extent.  They  may,  however,  be  limited  by 
the  directrices  or  in  other  ways.  A  definitely  limited  surface 
sometimes  can  be  shown  in  projection  by  projecting  its  outline, 
as  in  the  circular  stairs  shown  in  Fig.  178.  In  most  cases, 
however,  a  better  conception  is  given  if  some  of  the  elements 


186 


DESCRIPTIVE   GEOMETRY 


[YIII,  §  162 


of  the  surface  are  drawn.  When  this  is  done,  as  in  Fig.  176, 
the  drawing  shows  the  modehng  of  the  surface  to  a  certain 
extent.  Another  possible  method  of  representing  a  warped 
surface  is  given  in  §  219. 

163.   The  Hyperboloid  of  Revolution.     I.   General  Proper- 
ties.    This  surface,  which  is  the  only  possible  warped  surface 


of  revolution  (§  161),  is  generated  by  revolving  a  straight  line 
about  an  axis  not  in  its  own  plane. 

As  in  the  case  of  every  surface  of  revolution,  any  section 
perpendicular  to  the  axis  will  be  a  circle.  That  point  of  the 
generatrix  which  is  nearest  the  axis  will  generate  the  smallest 
possible  circular  section,  called  the  circle  of  the  gorge.  From 
the  gorge,  the  surface  swells  out  as  the  generatrix  recedes  from 
the  axis.     The  general  form  of  the  surface  is  shown  in  Fig.  176. 


VIII,  §  163]  CURVED  SURFACES  187 

Sections  made  by  passing  planes  through  the  axis  are  hyper- 
bolas. Hence  the  same  sm'face  might  be  generated  by  an 
hyperbola  revolving  about  its  axis.  (Compare  with  the  parab- 
oloid, §  167.) 

II.  To  Draw  the  Projections  of  an  Hyperboloid  of 
Revolution.  In  Fig.  176,  let  us  suppose  given,  the  axis  and 
one  position  of  the  generatrix,  as  ab.  A  perpendicular  from 
the  axis  to  the  generatrix,  as  gh,  establishes  the  circle  of  the 
gorge  on  plan,  while  the  distances,  ga,  gh,  establish  the  radii  of 
the  bases.  The  projecting  planes  of  all  the  elements  will  be 
tangent  to  the  projecting  cylinder  of  the  gorge  circle.  Hence 
the  H  projection  of  every  element  will  be  tangent  to  the  gorge, 
as  a^b'',  c^d^,  etc.  The  V  projections  are  found  by  projecting  a^f 
c\  e'\  etc.,  to  the  upper  limiting  plane,  and  6\  d^J\  etc.,  to  the 
lower  plane  as  shown,  and  then  connecting  the  proper  points. 

III.  Double  Ruling.  It  will  be  noticed  that  the  element 
ab  and  the  line  jk  have  the  same  //  projection.  These  lines  lie 
in  the  same  vertical  plane,  are  inclined  at  the  same  angle  toward 
H,  and  intersect  at  o.  Corresponding  points  on  these  lines  are 
symmetrically  situated  with  respect  to  the  axis.  Hence  the 
same  surface  will  result  from  the  rotation  of  either  line.  A 
surface  that  may  be  thus  generated  by  the  motion  of  either 
of  two  generatrices  is  said  to  be  double-ruled.      (See  also  §  164.) 

IV.  To  Locate  a  Point  on  the  Surface.  Let  the  H  or  V 
projection  of  the  required  point  be  assumed  at  will,  within  the 
limits  of  the  surface.  If  the  H  projection  is  assumed,  pass  an 
element  through  the  assumed  projection  tangent  to  the  gorge. 
Next  determine  the  V  projection  of  the  element,  and  then  project 
the  point  from  the  plan  up  to  this  line. 

If  the  V  projection  of  the  point  is  assumed,  pass  an  H  plane 
through  the  assumed  projection.  This  plane  will  cut  a  circle 
from  the  surface.  This  circle  will  appear  as  a  line  on  V,  Find 
the  projection  of  this  circle  on  H,  and  project  down  to  it  from 
the  assumed  V  projection  of  the  point. 


188 


DESCRIPTIVE   GEOMETRY  \yni,  §  164 


164.  The  Hyperbolic  Paraboloid.  This  surface  is  generated 
by  a  line  that  slides  along  two  straight  directrices  not  in  the 
same  plane,  and  remains  parallel  to  a  plane  director.  It  is 
represented  in  Fig.  177,  as  limited  by  its  directrices,  the  lines 
ab  and  cd.     The  plane  H  is  the  plane  director. 

Successive  elements  of  the  surface  lie  in  successive  planes 
parallel  to  H,  and  these  planes  divide  the  directrices  pro- 
portionally, as  at  1,  2,  3  and  at  c,f,  g,  etc.     Hence  lines  joining 


1-f,  2-/,  Z-g,  etc.,  are  elements  of  the  surface.  It  will  be  evident 
that  if  any  two  lines  are  divided  into  the  same  number  of  equal 
parts,  and  if  the  points  thus  determined  are  connected  in  order, 
the  lines  thus  determined  will  be  parallel  to  some  plane.  Hence 
these  lines  will  be  elements  of  an  hyperbolic  paraboloid. 

In  Fig.  177  6,  the  lines  ac  and  hd  of  Fig.  177  a  are  redrawn. 
In  Fig.  177  a,  these  lines  were  elements  of  the  surface.  In 
Fig.  177  h,  the  same  lines  are  divided  proportionally  at  1,  3,  5 


VIII,  §  165] 


CURVED   SURFACES 


189 


and  at  2,  4,  6.  The  lines  1-2,  3-4,  etc.,  are  elements  of  an 
hyperbolic  paraboloid,  as  pointed  out  in  the  previous  paragraph. 
By  drawing  the  lines  I'e',  2'j' ,  etc.,  corresponding  to  \  e,  2/, 
etc.,  in  Fig.  177  a,  it  can  be  shown  that  the  points  x,  y,  etc.,  lie 
on  both  sets  of  lines.  (Let  the  student  complete  the  proof.) 
Thus  either  set  of  lines  gives  elements  belonging  to  the  same 
surface.     Hence  the  surface  is  a  double-ruled  surface  (§  163). 


Circular    stairs. 
(Helicoid  ) 


Arch  in  a  cylindrical  well. 
( Conoid ) 


PlAH 


Hood  for  a  small 
fireplace. 
(Conoid.) 


SKew  arch. 
(Corne  de  Vache) 


Fig.  17^ 


The  recessed   Marseilles  Gate, 
(Irregular.) 


165.  Other  Warped  Surfaces.  The  two  surfaces  described 
above  are  interesting  chiefly  because  they  illustrate  so  well  the 
characteristics  of  warped  surfaces  in  general.  In  practice  they 
are  rarely  encountered.  Some  of  the  forms  that  are  niore  usual 
in  construction  work,  and  which  bear  some  resemblance  to 
cones,  cylinders,  etc.,  are  shown  in  Fig.  178.  The  cotzozc?  forms 
have  one  straight  and  one  curved  directrix  and  a  plane  director. 


190  DESCRIPTIVE   GEOMETRY         \yiU,  §  165 

The  helicoid  which  is  shown  in  the  figure  is  the  one  known  as 
the  right  helicoid:  it  has  a  heUx  and  its  axis  as  its  directrices 
and  a  plane  H  as  its  director,  to  which  all  the  elements  are 
parallel.  Obhque  heUcoids  have  the  same  directrices,  but 
instead  of  a  plane,  they  have  a  conical  director. 

The  Cornede  Vache  (Fig.  178)  has  for  its  directrices  two  equal 
circles  in  parallel  planes,  and  a  straight  line  perpendicular  to  the 
planes  of  the  circles  and  passing  through  the  center  point  of 
a  line  joining  the  centers  of  the  circles.  The  generatrix  moves 
in  contact  with  these  three  lines. 

In  the  Marseilles  gate  (Fig.  178),  the  surface  over  the  door, 

between  the  elements  ac  and  bd,  is  generated  by  a  straight  line 

moving  on  the  curves  ab  and  cd,  and  along  the  straight  line  xy. 

The  surface  between  ac  and  e  has  for  directrices  the  curves 

e  and  ce  and  the  line  xy.     The  two  surfaces  have  a  common 

element  in  ac. 

EXERCISE    SHEET   LXIX 

1.  Given  the  lines  a  (66,  0,  12),  b  (66,  22,  12)  and  c  (56,  22,  15), 
d  (76,  0,  15).  Draw  the  hyperboloid  of  revolution  generated  by  rotat- 
ing cd  about  ab.  Locate  the  points  e  and  /,  which  are  on  the  surface 
just  drawn  and  are  2  from  H  and  14  from  V. 

2.  Using  the  lines  g  (49,  21,  20),  h  (30,  4,  2)  and  i  (28,  21,  25), 
j  (50,  4,  3)  as  directrices,  and  fl"  as  a  plane  director,  draw  an  hj^jerbohc 
paraboloid. 

3.  Using  the  lines  g'  (24,  21,  20),  i'  (3,  21,  25),  and  f  (25,  4,  3), 
h'  (5,  4,  2)  as  directrices,  retrace  the  surface  of  exercise  2.  Find  the 
point  m,  in  which  the  Une  k  (21,  24.  14),  I  (4,  12,  23)  pierces  the  surface. 

EXERCISE   SHEET   LXX 

1.  Draw  the  projections  of  the  warped  surface  which  has  three 
linear  directrices  as  follows:  (a)  a  straight  Hne  parallel  to  HA  and 
passing  through  the  point  a  (49,  0,  10) ;  (b)  a  semi-ellipse  lying  in  a 
plane  perpendicular  to  HA.  and  ha\'ing  its  center  at  a,  the  minor  axis 
being  If"  long  and  lying  in  H,  while  the  semi-major  axis  is  If"  long; 
(c)  a  semi-circle  1|"  in  diameter,  in  a  plane  perpendicular  to  HA  and 
with  its  center  at  b  (60,  0,  10). 

2.  Imagine  a  right  circular  cyhnder  2^"  long  with  its  axis  perpen- 
dicular to  V.     Let  its  diameter  be  1|",  and  the  center  of  the  rear  base 


VIII,  §  166]  CURVED   SURFACES  191 

be  at  c  (29,  15,  4).  Now  let  the  front  base  be  moved  downward  and 
to  the  right  till  its  center  is  at  d  (12,  6,  24).  During  this  motion  of 
the  front  base  let  it  be  rotated  in  its  own  plane,  in  a  clockwise  direction, 
through  120°;  the  rear  base  meanwhile  remaining  fixed.  During  the 
above  movements,  let  each  element  remain  straight  and  always  con- 
necting the  same  points  on  the  circular  bases,  but  varying  in  length. 
Draw  the  projections  of  the  resulting  warped  surface. 

EXERCISE   SHEET  LXXI 

Place  the  sheet  vertically,  with  HA  in  the  center.  Draw  the  pro- 
jections of  an  hyperboloid  of  revolution  whose  axis  is  the  line  a  (29, 
0,  20),  b  (29,  26,  20),  and  whose  generatrix  is  the  line  c  (46,  26,  25), 
d  (12,  0,  25).  Draw  also  the  projections  of  a  conoid  whose  directrices 
are  the  line  e  (46,  0,  8),  /  (46,  17,  8),  and  a  semicircle  4j"  in  diameter 
with  its  center  at  g  (14,  0,  24).  The  plane  of  the  semicircle  is  per- 
pendicular to  e^  and  H  is  used  as  a  plane  director.  Find  the  line  of 
intersection  of  the  given  surfaces. 

166.  Double-Curved  Surfaces.  The  chief  characteristics  of 
double-curved  surfaces  are  given  in  §§  138-142.  They  are  as 
follows:  (1)  The  generatrix  is  a  curve.  ^  (2)  Such  a  surface  is 
not  covered  by  straight  lines.  (See  p.  156.)  (3)  These  surfaces 
may  be  generated  either  by  transposition  or  by  revolution. 

By  far  the  most  important  of  the  double-curved  surfaces 
are  those  which  are  generated  by  revolution.  The  simplest 
case  is  that  of  the  sphere,  formed  by  rotating  a  circle  on  its 
diameter.  In  the  same  class  are  balusters,  vases,  and  most  of 
the  products  of  the  lathe,  the  potter's  wheel,  and  the  metal 
spinner,  excepting  only  such  as  take  the  form  of  a  cone,  cylinder, 
or  hyperboloid  of  revolution  (§  161). 

When  a  curved  generatrix  follows  a  curved  directrix  with 
a  motion  of  transposition,  a  double-curved  surface  is  formed. 
This  case  covers  a  wide  variety  of  surfaces,  not  all  of  which 
have  distinct  names.     An  example  is  shown  in  Fig.  186. 

1  The  case  of  surfaces  generated  by  a  line  which  changes  form  are  not 
considered  here.  Thus  a  cord  which  changes  position  while  in  vibration 
would  have  many  positions  in  which  it  would  be  straight,  while  in  most 
positions  it  would  be  curved.  A  surface  so  generated  would  contain 
many  straight  lines,  though  the  surface  as  a  whole  would  not  be  ruled. 


192  DESCRIPTHT:  geometry         [VIII,  §  167 

167.  Typical  Double-Curved  Surfaces  of  Revolution.  A 
circle,  when  rotated  on  a  diameter,  generates  a  sphere.  An 
ellipse,  rotated  on  its  diameter,  gives  a  spheroid.  AVhen  the 
major  axis  of  the  ellipse  is  used  as  the  axis  of  rotation,  the 
surface  generated  is  called  a  prolate  spheroid.  When  the  minor 
axis  is  used,  the  resulting  surface  is  an  oblate  spheroid. 

Paraboloids  and  hyperboloids  are  generated  in  a  similar 
manner  by  the  revolution  of  parabolas  or  hyperbolas.  One  of 
the  possible  hyperboloids  is  the  warped  surface  described  in  §  163. 

A  circle  which  revolves  about  an  axis  wholly  outside  itself 
generates  an  annular  torus,  the  typical  ring  form.  Other  forms 
of  the  torus  are  given  by  revolving  other  closed  curves.  The 
bases  of  classic  columns,  and  some  of  the  capitals,  contain  various 
forms  of  the  torus,  some  of  which  have  special  names. 

168.  Sections.  Every  point  on  the  generatrix  of  a  surface 
of  revolution  describes  a  circle,  the  plane  of  which  is  perpen- 
dicular to  the  axis  of  revolution.  Hence  any  plane  passed 
perpendicular  to  the  axis  will  cut  from  such  a  surface  a  circle, 
the  center  of  which  is  on  the  axis.  This  fact  gives  a  basis  for 
the  solution  of  many  of  the  problems  which  arise  in  connection 
with  double-curved  surfaces.  Any  plane  passed  through  the 
axis  of  a  double-curved  surface  of  revolution  cuts  the  surface  in 
a  curve  identical  with  the  generatrix.  Such  a  plane  is  called  a 
meridian  plane ;  the  section  is  called  a  meridian  section. 

169.  Representation.  Normally,  double-curved  surfaces  are 
drawn  with  the  axis  parallel  to  one  of  the  planes  of  projection. 
When  in  this  position,  one  projection  consists  of  a  meridian 
section,  while  the  other  is  a  circle  representing  the  largest 
diameter,  together  with  other  circles  showing  the  size  of  the 
surface  at  sections  where  the  curvature  changes  (Fig.  179). 

The  projections  of  a  sphere  consist  of  two  equal  circles.  The 
plan  shows  the  upper  half  of  the  sphere  as  visible,  while  the 
elevation  shows  the  forward  half  as  visible.  Onl^'  the  upper 
front  quarter  of  the  surface  is  shown  as  visible  on  both  projec- 


VIII,  §  170] 


CURVED   SURFACES 


193 


tions,  and  the  lower  rear  quarter  is  not  visible  on  either.  This 
is  typical  of  most  double-curved  surfaces  of  revolution.  After 
sufficient  practice,  the  beginner  will  be  able  to  distinguish  clearly 
between  visible  and  invisible  points  on  a  surface. 

170.  To  Locate  a  Point  on  the  Double-Curved  Surface  of 
Revolution.  In  attempting  to  locate  a  point  on  a  plane  surface 
(§  65),  and  again  on  a  ruled  sur- 
face (§  148),  it  was  found  neces- 
sary to  first  determine  a  line  on 
the  surface.  The  same  necessity 
will  be  found  in  the  present 
problem.  In  this  case  the  line 
most  often  used  is  a  circle, 
since  that  is  the  simplest  line 
that  can  be  drawn  on  a  double- 
curved  surface. 

In  Fig.  179  are  shown  the 
projections  of  a  prolate  sphe- 
roid. Let  it  be  required  to 
locate  a  point  on  the  surface. 
One  projection  of  the  required 
point  may  be  assumed  at  will, 
within  the  given  limits  of  the 
surface.  Let  x'"  be  assumed,  as 
shown.  The  point  x  lies  in  the 
horizontal  plane  P.     If  it  is  to 

lie  on  the  spheroid,  it  must  be  on  the  line  of  intersection  be- 
tween the  spheroid  and  P.  This  line  of  intersection  is  seen  as 
a  circle  on  plan,  the  diameter  of  which  is  ah.  The  H  projection 
of  X  lies  on  this  circle  and  below  .t^,  i.e.  either  at  x^  (on  the  front 
half)  or  at  x'^  (on  the  rear).  Again  the  assumed  projection  may 
be  chosen  on  plan,  as  at  y^.  The  point  y  lies  on  the  surface 
circle  y^m,  which  can  be  located  on  elevation  either  in  the  plane  M 
or  A^     The  V  projection  of  y  will  be  found  either  at  y"  or  y'"". 


Fig.  179 


194 


DESCRIPTIVE  GEOMETRY 


[VIII,  §  171 


"  171.  To  Draw  a  Line  Tangent  to  a  Double-Curved  Surface  of 
Revolution  at  a  Given  Point  on  the  Surface.  If  a  plane  cuts 
from  the  given  surface  any  curve,  any  line  in  the  cutting  plane 
and  tangent  to  the  curve  is  tangent  to  the  surface  (Fig.  153). 

In  Fig.  180,  let  it  be  required  to  pass  a  line  tangent  to  the 
paraboloid  at  the  point  x.     First  let  a  meridian  plane  M  be 


Fig.  ISO 


passed  through  x.  This  plane  will  cut  a  parabola  from  the 
surface.  The  projection  of  this  parabola  could  be  determined 
by  the  method  of  §  149,  and  the  tangent  could  be  drawn  at 
once. 

It  will  be  found  easier,  however,  to  rotate  the  entire  con- 
struction (toward  the  right,  in  the  figure)  on  the  axis  ah,  until 
the  cutting  plane,   M,  is  parallel  to    V .     The  curve  of  inter- 


VIII,  §  173]  CURVED   SURFACES  195 

section  between  the  paraboloid  and  the  plane  M  will  then 
appear  as  the  meridian  section  6^.c'^£^^  and  the  point  x  will 
move  to  x'. 

The  line  c'x\  drawn  tangent  to  the  meridian  curve,  is  in  the 
same  plane  with  the  curve.  Hence  it  is  tangent  to  the  parabo- 
loid. 

If  now  the  cutting  plane  is  rotated  back  to  its  original  posi- 
tion, the  point  c'  will  move  to  c  and  x'  to  x,  the  tangent  line 
being  ex.  The  lines  ex  and  e'x'  are  elements  of  a  right  circular 
cone  whose  apex  is  at  o,  and  whose  base  passes  through  x. 

The  Hne  just  determined  does  not  constitute  the  only  possible 
solution  to  the  problem.  Many  lines  may  be  drawn  tangent 
to  such  a  surface  at  any  given  point.  Any  cutting  plane 
whatever  through  the  given  point  would  yield  a  curve  of  inter- 
section to  which  a  tangent  line  could  be  drawn.  Perhaps  the 
easiest  solution  for  this  problem  results  from  the  use  of  a  hori- 
zontal cutting  plane,  since  such  a  plane  gives  a  circular  line  of 
intersection.  In  general,  the  student  should  be  familiar  with 
the  solution  using  either  a  right  section  ( H  cutting  plane)  or 
a  meridian  section,  as  above. 

172.  To  Pass  a  Plane  Tangent  to  a  Double-Curved  Surface  of 
Revolution,  at  a  Given  Point  on  the  Surface.  Draw  any  two 
tangent  lines  through  the  given  point,  as  explained  in  §  171. 
These  lines  will  determine  the  required  tangent  plane. 

In  the  special  case  of  the  sphere,  we  may  draw  the  radius 
terminating  at  the  given  point  and  we  may  pass  a  plane  per- 
pendicular to  this  radius  and  through  its  extremity  (§  82). 
Such  a  plane  will  be  tangent  to  the  sphere. 

173.  To  Draw  a  Right  Circular  Cone  with  a  Given  Apex  and 
Tangent  to  a  Given  Sphere.  The  axis  of  the  cone  will  pass 
through  the  center  of  the  sphere,  and  the  line  of  contact  will 
be  a  circle.  The  simplest  possible  view  of  the  surfaces  will 
be  that  one  which  is  projected  on  a  plane  parallel  to  the  axis 
of  the  cone. 


196 


DESCRIPTIVE   GEOMETRY 


[VIII,  §  173 


r  / 


In  Fig.  181,  let  the  sphere  whose  center  is  c,  and  the  apex 
of  the  cone,  o,  be  given.  Find  the  projections  of  o  and  c  on  a 
V  plane  taken  parallel  to  oc.     These  are  shown  at  o"'  and  c''. 

The  elements  of  sight  of  the  cone 
as  seen  on  this  new  plane  will  be 
tangent  to  the  element  of  sight 
of  the  sphere,  and  the  circle  of 
contact  between  the  surfaces  will 
appear  as  the  straight  line  a'^'b^' 

It  is  now  necessary  to  establish 
the  circle  of  contact  on  the  H  and 
T"  projections  of 
the  sphere,  which  " 
will  appear  as 
ellipses.  On  the 
H  projection,  the 
minor  axis  will  be 
projected  in  a  line 
parallel  to  V,  at 
b^a''.  The  major 
axis  is  perpendicu- 
lar to  ¥a^,  at  its 
center,  and  its 
length  is  equal  to 
a^'b^'  as  given  on 
the  V  projection. 
The  H  projection 
of  the  circle  of  con- 
tact can  be  com- 
pleted by  drawing 

an  ellipse  through  a^d^'b^e^.  The  V  projection  can  be  estab- 
lished point  by  point  from  the  H  and  V  projections,  or  by 
determining  the  V  projections  of  the  axes  ab  and  de,  which 
will  be  conj\igcite  diameters  of  the  required  ellipse  (§  121). 


Fig.  181 


VIII,  §  174] 


CURVED  SURFACES 


197 


The  elements  of  sight  of  the  cone  can  now  be  drawn  through 
o  and  tangent  to  the  circle  of  contact.  If  carefully  drawn,  the 
elements  of  sight,  the  circle  of  contact,  and  the  projection  of 
the  sphere  will  pass  through  a  common  point. 

174.  To  Draw  the  Projections  of  a  Double-Curved  Surface  of 
Revolution  in  any  Position.  I.  The  Envelope  Method. 
When  a  surface  of  revolution  is  placed  so  that  its  axis  is  parallel 
to  a  plane  of  projection,  one  line  of  sight  is  a  meridian  section, 
while  the  other  is  a  right  section  (§  169).  If  the  axis  is  in- 
clined to  the  planes  of  projection,  the  lines  of  sight  shift  to  new 
positions,  which  are  not  easily  determinable.  The  method  used 
below  to  determine 
the  lines  of  sight  is 
called  the  envelope 
method.  (See  §  142 
and  Fig.  154.)  This 
method  is  used  in 
many  of  the  more 
complex  problems  in 
shades  and  shadows. 

Figure  182  shows  a 
spheroid  with  its  axis 
inclined  toward  H  and 
V.  A  right  section, 
as  ab,  is  projected  on  H  as  the  ellipse  a^b^.  The  lines  ea  and 
fb  are  passed  tangent  to  the  spheroid,  perpendicular  to  H, 
and  touching  the  right  section  at  a  and  b.  The  points  a  and 
b  are  on  the  line  of  sight  of  the  spheroid  and  a''  and  b^  are 
points  on  the  H  projections  of  the  spheroid.  Now  let  the 
spheroid  be  considered  as  a  surface  of  transposition  (§  139), 
generated  by  a  variable  right  section,  ab.  Successive  positions 
of  the  right  section  give  ellipses  as  their  H  projections. 
Successive  positions  of  the  projectors  ca  and  fb  generate  a 
cylinder  (or  an  envelope)  which    incloses  the  spheroid.     The 


Fig.   182 


198 


DESCRIPTR  E   GEOMETRY  [VIII,  §  174 


intersection  of  the  envelope  with  H  is  a  curve,  tangent  to  the 
successive  projections  of  the  right  section,  which  determines 
the  projection  of  the  surface. 


Fig.  183 


VIII,  §  175]  CURVED   SURFACES  199 

II.  The  General  Method.  In  Fig.  183,  let  it  be  required 
to  draw  the  projections  of  the  surface  shown  at  (a)  when  its 
axis  is  incHned  to  H  and  V  as  indicated  by  the  hne  he. 

The  general  method  which  will  be  followed  ^  consists  in  draw- 
ing the  projection  of  the  given  surface  on  a  V  plane,  parallel 
to  the  given  position  of  the  axis,  as  shown  at  ih)  in  the  figure ; 
and  then  using  this  projection  as  a  basis  to  construct  the  re- 
quired H  and  V  projections. 

In  the  figure,  the  line  V2''  is  the  V  projection  of  a  circular 
right  section  of  the  surface.  The  //  and  V  projections  of  this 
circle  may  be  determined  by  the  method  of  §  128,  as  shown 
at  1''4''  and  1^4^  For  any  other  section,  as  5-6,  and  for  as 
many  others  as  may  be  desired,  the  projections  may  be  found 
in  the  same  manner  The  common  tangents  to  these  ellipses 
give  the  projections  of  the  required  surface. 

Where  any  portion  of  the  surface  is  a  true  cone  or  cylinder 
(as  between  .1  and  J5,  C  and  D,  in  the  figure),  it  is  only  necessary 
to  determine  the  limiting  sections,  the  tangents  between  them 
being  straight  lines.  The  line  of  sight  of  one  portion  of  the 
surface  may  fall  within  the  projection  of  another  part,  as  at 
Z.  This  merely  means  that  the  surface  here  resembles  a  cone, 
the  projection  of  the  apex  falling  within  that  of  the  base. 

[Note.  In  attempting  the  solution  of  such  a  problem  as  the  above, 
the  student  should  find  both  the  projections  of  a  given  right  section 
before  attempting  to  determine  anything  with  regard  to  the  next  one. 
A  common  tendency  is  to  determine  aU  the  H  projections  first.  This 
usually  leads  to  endless  confusion  and  mistakes.] 

175.  To  Find  the  Line  of  Intersection  between  a  Plane  and 
any  Surface  of  Revolution.  The  general  considerations  apply- 
ing to  the  intersection  of  surfaces  (§  149)  suffice  for  this  case. 
Since  all  right  sections  of  the  surface  are  circles,  the  auxiliary 
planes  used  will  be  those  that  will  cut  right  sections. 

1  Tliis  method  is  identical  in  principle  with  that  of  §  53  (for  prisms)  and 
§  146,  II  (for  cones). 


200 


DESCRIPTIVE   GEOMETRY 


[VIII,  §  175 


Figure  184  shows  the  intersection  between  a  plane  A',  and  an 
annular  torus.  It  will  be  noticed  that  there  are  points  of 
tangency  between  the  lines  of  sight  of  the  torus  and  the  curve 
of  intersection.  Any  such  curve  of  intersection  is  usually 
tangent  to  the  sight  lines  in  two  points  or  more. 

The  special  case 
of  the  preceding 
problem  which  in- 
volves a  cutting 
plane  perpendicu- 
lar to  H  or  V  is 
X^  frequently  used  in 
later  problems. 

176.  To  Find  the 
Points  in  which  a 
Line     Tierces     a 
Surface  of  Revolu- 
tion.    This   prob- 
lem admits  of  the 
same  solution  that 
was  used  in  §  151, 
I.     The   auxiliary 
plane  will  usually 
be  chosen  through 
the  line  and  per- 
pendicular to  H  or  V.     The  intersection  of  this  plane  with  the 
given  surface  can  be  found  as  in  §  175.     The  points  where  this 
curve  meets  the  given  line  are  the  required  piercing  points. 

177.  To  Find  the  Line  of  Intersection  between  any  Two  Sur- 
faces of  Revolution.  (See  also  §  149.)  The  methods  given 
below  will  apply  to  either  double-curved  or  ruled  surfaces  of 
revolution.  Three  cases  may  be  distinguished,  depending 
on  the  relations  that  exist  between  the  axes  of  the  given 
surfaces. 


VIII,  §  177] 


o 


URVED   SURFACES 


201 


I.  Parallel  Axes.     In  this  case  planes  perpendicular  to 
the  axes  will  cut  circles  from  each  of  the  surfaces. 

II.  IxTERSECTixG  AxES.  It  is  often  difficult  to  find  a  plane 
which  cuts  l)oth  surfaces  in  curves  which  are  sufficient!}'  simple 
to  give  an  easy  solu- 
tion. A  simple  solu- 
tion results,  however, 
if  spheres  are  uecd 
instead  of  planes  for 
cutting  the  surfaces. 

If  any  surface  of 
revolution  is  cut  by 
a  sphere,  the  center 
of  which  lies  on  the 
axis  of  the  surface, 
the  line  of  inter- 
section is  a  circle. 
Therefore,  if  two  sur- 
faces of  revolution 
have  axes  that  in- 
tersect, the  point  of 
intersection  may  be 
used  as  the  center  of 
a  series  of  cutting 
spheres.  Each  of 
these  spheres  will  cut 
a  circle  from  each  of 
the  surfaces  and  the 

intersections  of  these  circles  give  points  on  the  required  line  of 
intersection. 

In  Fig.  185,  let  the  vase  form  and  cone  of  revolution  be  given, 
and  let  the  axes  intersect  at  o.  \Yith  o  as  a  center,  generate 
a  sphere,  as  shown  by  abed.  This  sphere  will  cut  from  the  cone 
a  circle  which  is  shovv'n  on  elevation  at  a'-'C.     From  the  vase 


Fig.  185 


202 


DESCRIPTIVE   GEOMETRY 


[VIII,  §  177 


it  will  cut  a  circle  shown  on  elevation  at  b^'d''.  These  circles 
intersect  at  two  points  x  and  x\  which  coincide  on  the  elevation. 
On  plan,  the  circle  cut  from  the  vase  is  shown  at  x^'^b^x^d^.     The 

points  on  the  required  line  of 
intersection,  x  and  x\  are  now 
fully  determined.  Other  spheres, 
concentric  vrith  tlie  one  just 
drawn,  will  yield  other  points  on 
the  required  curve. 

III.  Axes  not  Ixtersectixg 
AND  NOT  Parallel.  In  this 
case,  it  is  not  possible  to  state 
any  solution  that  will  be  best  for 
all  cases.  Any  cutting  surface 
which  gives  a  simple  intersection 
with  one  surface  will  probably 
give  a  less  simple  one  on  the 
other  surface,  rendering  the  solu- 
tion tedious,  but  a  solution  is 
always  possible  according  to  the 
general    principles    explained    in 

§  l-t9. 

178.  To  Draw  the  Projections 
of  the  Serpentine.  The  serpen- 
tine is  a  surface  generated  by  a 
sphere  which  moves  so  that  its 
center  follows  a  given  helical 
directrix. 

Figure  186  shows  the  projec- 
tions of  the  serpentine  when 
the  axis  of  the  helix  is  perpendicular  to  H.  Various 
positions  of  the  spherical  generatrix  are  drawn.  These 
are  connected  by  a  common  tangent  as  explained  in 
§  174. 


Fig.   186 


VIII,  §  178]  CURVED  SURFACES  203 


EXERCISE    SHEET   LXXII 

1.  The  point  a  (66,  12,  12)  is  the  center  of  a  sphere  2j"  in  diameter. 
Locate  the  point  h,  which  lies  on  the  surface  of  the  sphere,  2|"  above 
H  and  2"  in  front  of  V.  Through  b  draw  a  diameter  of  the  sphere. 
Call  its  opposite  end  c.  Through  h  draw  a  line  tangent  to  the  sphere. 
Through  c  pass  a  plane  tangent  to  the  sphere. 

2.  The  point  /  (40,  12,  10)  is  the  center  of  a  prolate  spheroid,  whose 
major  axis  is  vertical  and  3"  long.  Find  the  line  of  intersection  be- 
tween the  given  surface  and  the  plane  Z  (51,  Z^  60°  r,  Z"  60°  r). 

3.  A  circular  arc  passes  through  the  points  g  (13,  22,  9),  h  (19,  11, 
9),  and  y  (13,  0,  9).  Let  this  arc  be  rotated  on  gj  as  an  axis,  forming 
a  double-curved  surface  of  revolution.  Find  where  the  Hne  k  (5,  20, 
4),  I  (22,  4,  13)  intersects  the  surface. 


EXERCISE    SHEET   LXXIII 

Place  the  sheet  vertically.  Take  HA  4"  below  the  top  border  line. 
Draw  the  projections  of  the  vase  form  shown  in  Fig.  208,  using  as  an 
axis  the  line  a  (44,  24,  32),  b  (26,  6,  2).  Let  the  length  of  the  vase  be 
about  3". 

EXERCISE   SHEET   LXXIV 

1.  Given  the  annual  torus  whose  inside  diameter  is  1|"  and  whose 
outside  diameter  is  Si".  Let  the  center  be  at  a  (64,  12,  14),  and  let 
the  plane  of  the  top  be  horizontal.  Given  also  a  right  circular  cyHnder, 
2|"  in  diameter,  resting  on  H,  with  the  center  of  its  base  at  b  (60,  0,  16). 
Find  the  line  of  intersection. 

2.  Given  an  annular  torus  described  in  (1)  above.  Let  the  center 
be  at  c  (23,  12,  14).  Given  also  a  right  circular  cone,  3"  in  diameter 
and  3^"  high.  Let  the  center  point  of  the  axis  of  the  cone  be  at  c, 
and  let  the  axis  be  parallel  to  V  and  inclined  at  an  angle  of  30°  to  H. 
Find  the  line  of  intersection. 


CHAPTER   IX 
APPLICATIONS   BASED    ON    CHAPTERS   I   TO   V  i 

179.  Introduction.  The  conditions  of  a  concrete  problem 
are  often  more  easily  seen,  and  its  solution  is  more  readily  per- 
formed, than  are  those  of  an  abstract  problem  of  a  similar 
nature.  For  this  reason,  one  is  always  tempted,  when  commenc- 
ing problems  similar  to  those  in  this  and  in  the  following  chapters, 
to  discard  the  fundamental  ideas  on  which  the  solutions  are 
based,  to  proceed  with  more  or  less  vague  and  disjointed 
methods,  and  to  be  satisfied  with  intuitive  solutions.  These 
soon  become  set  rules  or  formulas.  When  this  happens,  the 
time  spent  on  such  problems  is  worse  than  lost.  If,  however, 
the  student  seeks  to  trace  back  each  step  in  the  solution  of  a 
concrete  problem  to  its  underlying  geometric  principle,  and  to 
build  his  work  on  perfectly  general  laws,  much  benefit  may  be 
derived  from  solving  problems  of  this  nature. 

180.  Simple  Plan  and  Elevation.  (§§  1  to  25.)  In  starting 
to  lay  out  the  plan  and  elevations  of  a  simple  building,  such 
as  that  shown  in  Fig.  7,  it  is  common  practice  to  develop  the 
plan  quite  separately  from  the  elevations,  often  on  different 
drawing  boards. 

In  order  to  keep  the  plan  and  elevations  in  agreement  some 
common  basis  of  reference  is  needed.  For  this  purpose  vertical 
planes  are  used  that  cut  the  building  on  its  principal  axes.  These, 
seen  on  plan,  give  the  main  axis  lines.  (See  Figs.  197,  211.) 
From  their  relations  to  these  planes  the  positions  of  all  points 
can  be  shown  on  plan. 

1  Exercise  sheets  to  ucconipany  this  chapter  arc  given  on  page  295. 

204 


IX,  §  180]  APPLICATIONS  205 

Having  correctly  located  the  main  points,  say  the  four  out- 
side corners,  other  points  can  be  located  on  the  plan  from  their 
known  relations  to  these  points  or  to  the  axes. 

For  the  purpose  of  locating  heights  (drawing  elevations), 
a  horizontal  plane  is  used.  This  is  generally  taken  at  the  level 
of  the  first  floor,  or  at  some  other  level  arbitrarily  chosen. 
Such  a  plane  is  called  a  datum  plane.  It  is  usually  established 
by  some  permanent  natural  feature  of  the  land,  such  as  the  level 
of  the  lake  in  Fig.  247,  or  by  some  artificially  established  mark, 
such  as  the  top  of  the  curbing  in  a  city  street.  The  datum 
plane  is  shown  on  the  elevations  as  a  horizontal  line,  called  the 
datum  line.  This  datum  line  is  the  first  line  to  be  drawn  on  an 
elevation.  All  heights  are  measured  from  it,  either  upward  or 
downward. 

Next,  in  order  to  relate  the  plan  and  the  elevations,  an  axis 
line  is  drawn.  This  line  should  represent  one  of  the  axis  planes 
used  in  drawing  the  plan.  From  this  line  all  horizontal  lengths 
are  measured,  either  toward  the  right  or  left.  The  correct 
development  of  the  drawings  is  thus  assured. 

It  is  usually  wise  to  start  with  the  plan.  Develop  it  as  far 
as  is  easily  possible.  Then  pass  to  the  elevation.  Possibly 
this  cannot  be  completed  at  once.  Perhaps  now  the  plan 
can  be  completed.  Passing  thus  from  one  drawing  to  the 
other,  the  whole  can  finally  be  finished.  The  order  in 
which  the  work  is  to  be  done  should  be  carefully  planned  in 
advance. 

As  any  plan  develops,  the  beginner  should  keep  in  mind 
which  lines  are  normal-case  lines  and  which  ones  are  special 
cases,  which  ones  show  true  lengths  and  inclinations,  and  which 
ones  are  foreshortened  in  projection. 

The  fact  that  the  planes  of  projection,  as  formally  erected  in 
problems  of  descriptive  geometry,  are  apparently  discarded, 
should  not  lead  to  a  different  conception  of  the  fundamental 
facts  involved.     The  floor  or  datum  level  is  in  fact  what  we 


206 


DESCRIPTIVE   GEO:\IETRY 


[IX,  §  180 


previously  called  the  H  plane.  The  axis  planes  are  in  fact  V 
or  P  planes. 

\Yhen  working  in  this  manner  (purely  as  a  matter  of  con- 
venience), the  planes  of  projection  are  not  so  opened  as  to  lie 
adjacent  to  the  same  axis  hue,  but  they  are  placed  at  will, 
perhaps  on  different  drawing  boards.  However,  the  fundamental 
relations  between  the  reference  planes  are  maintained.  The 
V  planes  passed  through  the  plan  axis  and  the  datum  plane 
for  levels  correspond  exactly  to  V,  P,  and  H,  as  used  in  the 
abstract  problem. 

181.  Sections.  Plans  and  elevations  are  sufficient  for  the 
description  of  simple  solids.     When  the  shape  of  the  sohd  to  be 


-SfCIIOII 
PUKE 


SECTION 


Fig.  187 

shown  is  such  as  to  involve  reentrant  angles  or  when  the  object 
is  hollow  and  it  is  necessary  to  show  the  thickness  of  the  shell 
or  the  interior  arrangement,  a  different  kind  of  drawing,  called 
a  section,  is  required. 

The  necessity  for  making  a  section  lies  in  the  fact  that  in 
ordinary  projections  certain  parts  of  the  object  are  concealed 
by  other  parts.  The  method  used  is  to  remove  the  obstructing 
parts. 

The  object,  a  section  of  which  is  to  be  drawn,  is  considered 
as  cut  by  a  plane,  along  the  desired  line  (Fig.  187).  One  part 
of  the  object  is  then  removed  and  the  remaining  part  is  shown 
as  projected  on  the  cutting  plane. 


IX,  §  181]  APPLICATIONS  207 

It  is  usual  to  indicate  all  solid  parts  which  are  cut  by  the 
section  plane  in  some  distinctive  manner.  Cross  hatching, 
soKd  black,  and  other  indications  are  commonly  used.  Dif- 
ferent indications  are  used  on  the  same  drawing  in  order  to 
show  different  materials. 

Section  planes  are  commonly  vertical  planes  so  chosen  as 
to  coincide  with  one  of  the  principal  axes  of  the  object,  although 
much  latitude  is  allowed  in  this  respect.  For  special  reasons 
section  planes  may  be  chosen  in  almost  any  position  in  order 
to  fully  explain  the  object  in  hand. 

x\ll  parts  of  the  object,  both  inside  and  outside,  which  lie 
beyond  the  section  plane  are  projected  on  the  section  plane 
in  the  usual  manner. 

When  an  object  has  been  drawn  in  plan  and  in  elevation, 
and  a  section  is  required,  it  is  necessary  first  to  determine  where 
the  section  plane  may  be  located  in  order  to  be  most  useful  in 
describing  the  hidden  parts.  This  having  been  decided,  the 
trace  of  the  plane  is  shown  on  plan.  All  horizontal  distances 
for  constructing  the  section  can  be  found  by  projecting  points 
back  to  this  line.  The  vertical  relations  are  established  from 
the  elevation.  The  section  is  thus  built  up  point  by  point  from 
plan  and  elevation  in  the  same  way  that  a  profile  projection  is 
determined  from  known  horizontal  and  vertical  projections. 

Floor  plans  of  a  building  constitute  a  special  case  of  sections. 
Here  the  section  plane  is  chosen  in  a  horizontal  position, 
cutting  the  walls  at  about  the  mid-height  of  doors  and  windows. 
The  parts  above  being  removed,  a  single  floor  is  shown  as  pro- 
jected on  plan,  the  walls  being  cross  hatched  as  in  a  section. 

In  making  sections  through  a  complicated  structure  it  is 
sometimes  necessary  to  use  cutting  planes  that  are  not  con- 
tinuous, one  part  being  out  of  line  with  another,  in  order  to 
explain  a  particular  feature  of  the  construction.  When  this 
is  done,  care  must  be  taken  to  avoid  overlapping  of  different 
parts  on  the  section  drawing. 


208 


DESCRIPTIVE  GEOMETRY 


[IX,  §  182 


182.  A  Corner  Rafter  (§§  27,  28).  Figure  188  shows  the 
corner  of  a  show  rafter  cornice.  It  will  be  noted  that  the  hip 
rafter  is  of  different  size  and  shape  from  the  others  but  that 
the  shapes  are  definitely  related.  Commonly  the  jack  rafter 
is  designed  first  and  from  it  the  shape  of  the  hip  rafter  is  de- 
termined. 

Any  horizontal  line  (as  A  A' A)  parallel  to  the  wall,  and 
touching  each  jack  rafter  at  a  given  point,  determines  the 
corresponding  point  on  the  corner  rafter. 


Fig.  188 


In  Fig.  189,  let  the  plan  (c),  together  with  the  section 
drawn  above  it,  be  given.  Let  the  section  (6)  show  the  shape 
desired  for  the  jack  rafters.  The  plan  shows  the  position  of 
both  the  jack  and  corner  rafters.  Let  it  be  required  to  determine 
the  proper  shape  for  the  corner  rafter. 

The  method  to  be  followed  consists  in  finding  the  projection 
of  the  rafter  on  a  vertical  plane  erected  parallel  to  it  (§  27), 
as  shown  by  H'A'. 

Lengths  are  projected  from  the  plan,  while  heights  can  be 
measured  from  any  datum  plane  (as  BB)  on  the  section.  The 
first  lines  determined  should  be  the  top  and  bottom.     Then 


X,  §  182] 


APPLICATIONS 


209 


the  ornamental  part  may  be  determined  by  locating  a  sufficient 
number  of  points  on  the  curve  to  give  it  the  same  character 
as  the  curve  from  which  it  is  derived. 


Fig.  189 


The  true  shape  of  the  corner  rafter  having  been  found  as  above, 
a  valuable  exercise  will  be  found  in  determining  the  front  ele- 
vation of  the  corner,  as  shown  at  (a). 


210 


DESCRIPTIVE  GEOMETRY 


[IX,  §  183 


183.  A  Pipe  Shaft  (§§  27,  31  to  3G,  4G,  53).  Figure  190 
shows  a  portion  of  the  floor  and  walls  of  a  building,  and  two 
points  a  and  h,  in  isometric  projection.  The  line  ah  is  to  be  the 
center  of  a  pipe  shaft,  3'  0"  square.  The  problem  is  to  locate 
the  holes  required  in  the  floor  and  wall,  and  to  show  the  shape 
of  each. 

In  Figure  191,  the  above  conditions  are  shown  in  plan  and 
elevation.     The  method  of  procedure  will  be  to  find  first  the 


Fig.  190 

lines  which  form  the  edges  of  the  shaft,  then  to  find  where  each 
of  these  lines  pierces  each  face  of  the  floor  and  walls,  and  finally 
to  connect  these  piercing  points  so  as  to  show  the  required 
holes. 

In  this  problem  it  will  be  wise  to  establish  no  formal  axis  line, 
but  to  use  the  front  and  rear  faces  of  the  wall  as  V  planes,  and 
the  ceiling  and  floor  as   //  planes. 

Let  the  line  ab,  Fig.  191,  represent  the  axis  of  the  shaft.  This 
shaft  is  to  have,  as  its  right  section,  a  square  3'  0"  on  each  side. 
The  projections  of  such  a  section,  having  its  center  at  any 
point,  o,  on  ah,  may  be  constructed  as  described  in  §  46. 


IX,  §  183] 


APPLICATIONS 


211 


Having  established  the  four  corners,  c,  d,  e,foi  such  a  section, 
the  edges  of  the  shaft,  eg  — fh,  can  be  drawn  parallel  to  ab. 


W     ^b" 


Fig.  191 


The  traces  of  these  lines  (§31)  on  the  surfaces  of  the  wall  and 
floor  establish  the  required  holes. 

[Note.  At  tliis  point  the  student  can  also  make  an  application  of 
the  principles  of  Chapters  I  to  V  to  the  elementary  problems  in  shades 
and  shadows,  given  in  §§  202  to  210.] 


CHAPTER  X 
APPLICATIONS   BASED    ON    CHAPTER  VI » 

184.  Roofing  the  Plan.  In  many  buildings  the  roof  treat- 
ment is  the  most  important  single  feature  of  the  exterior  design. 
The  form  depends  on  the  size  and  shape  of  the  plan.  Even  when 
the  plan  is  fixed,  considerable  latitude  of  selection  remains. 

The  elements  involved  are  usually  plane  surfaces,  set  at  a 
given  pitch,  meeting  so  as  to  shed  water.  The  simplest  case  is  a 
rectangular  plan  with  two  gables  shown  in  Fig.  192  a.  When 
the  width  of  the  plan,  a,  the  height  of  the  wall,  b,  and  the 
inclination  of  the  roof,  6,  are  given,  the  height  of  the  ridge,  c, 
becomes  fixed.  Any  three  of  these  dimensions  will  fix  the 
fourth.  Every  line  drawn  on  the  roof  parallel  to  the  eaves  is 
parallel  to  the  ground.  Any  other  line  on  the  roof  makes  an 
angle  with  the  ground  w^iich  may  vary  between  0°  and  0°. 

Figure  192  6  shows  a  typical  method  of  covering  a  square 
plan.  Here  the  eaves  are  carried  all  around  the  building. 
The  four  planes  meet  in  salient  angles  at  the  My  lines,  and  the 
ridge  shortens  to  a  iieak.  The  hip  lines  make  an  angle  w^ith 
the  ground  which  is  less  than  the  inclination  of  the  roof.  In  Fig. 
192  c,  the  same  plan  is  roofed  with  four  gables.  The  two  ridges 
are  the  same  height,  and  meet  at  a  point  of  crossing,  c,  while  the 
roof  planes  intersect  in  reentrant  angles  or  valleys.  The  slope 
of  the  valleys  is  the  same  as  that  of  the  hips  in  Fig.  192  b. 

If  the  point  of  crossing,  c,  in  Fig.  192  c,  be  raised  slowly,  the 
ridges,  ca,  cb,  begin  to  slope  downward  to  the  walls ;  the  in- 
clination of  the  valleys  becomes  greater ;  and  the  angle  between 
the  planes  meeting  on  the  valleys  is  changed.     The  roof  assumes 

1  Exercise  sheets  to  accompany  this  chapter  are  given  on  pages  295-299. 

212 


X,  §  184]        APPLICATIONS   OF   CHAPTER  VI 


213 


the  general  form  of  Fig.  192  d,  a  form  not  unusual  for  towers. 
If  c  is  raised  still  more,  at  a  certain  height  the  ridges  ca,  ch. 


Ridge. 


(c) 


Pea  If -^b 


(6)      Hipped  roof. 
c 


Fig.  192 

and  the  valley  cd  fall  in  the  same  plane,  giving  a  simpler  roof, 
as  shown  in  Fig.  192  e.  Determine  what  will  happen  if  the  point 
c  is  carried  still  higher. 

An  irregular  plan  like  that  shown  in  Fig.  192  /  lends  itself 
well  to  a  picturesque  treatment  of  the  fayades. 


214 


DESCRIPTIVE  GEOMETRY 


[X,  §  185 


Fig. 


185.  Intersection  of  Roof  Planes.  Figure  193  shows  two 
planes,  P  and  P^  each  inclined  to  H  at  the  same  angle,  6. 
These  planes  are  related  in  the  same  manner  as  two  roof  planes 

meeting  at  a  hip.  Let 
us  choose  a  point  o,  on 
the  line  of  intersection 
of  these  planes,  and  let 
0^  be  its  projection. 
Next  let  us  pass  planes 
through  00^  perpendicu- 
lar to  ah  and  6c,  giving 
the  triangles  oo^a  and 
oo^c.  These  triangles 
are  equal,  hence  o^a  is  equal  to  oV.  Then  the  triangles  o^ab  and 
o^cb  are  equal  and  the  angles  4>  and  4>'  are  equal.  Since  the 
angle  abc  is  cmy  angle,  it  may  be  stated  that  when  roofs  of  equal 
slope  intersect,  the  hip  or  valley,  as  seen  on  plan,  bisects  the 
angle  between  the  eaves.  Usually,  all  roofs  on  a  given  build- 
ing have  the  same  slope.  AVhen 
different  slopes  are  used,  the  plan 
of  the  hip  or  valley  does  not  bisect 
the  angle  between  the  walls. 

186.  Roofing  Plans.  In  drawing 
a  set  of  floor  plans  and  elevations, 
it  is  quite  easy  to  indicate  an  ar 
range ment  of  roof  planes  w^hich,  in 
^act,  cannot  exist,  or  that  would 
fail  to  shed  water,  or  that  would 
leave  some  portion  of  the  plan 
uncovered.  A  carefully  drawn  roof 
plan  will  reveal  any  such  fallacies. 

Figure  194  shows  a  case  in  point.  The  planes  a,  b,  e,  and  d  do 
not  meet,  either  in  a  ridge  or  in  a  peak,  but  leave  an  open  spot, 
e.     In  such  cases  a  (relatively)  flat  section  of  roof  is  sometimes 


Fig.  194 


X,  §  187]        APPLICATIONS   OF  CHAPTER  VI 


215 


5"(ribe) 


Roof  surface 

Fig.  196 


used.  The  visualization  of  a  roof 
plan  is  made  easier  by  a  shading 
Hke  that  in  Fig.  195. 

When  a  vertical  surface,  such  as 
the  side  of  a  chimney,  intersects  a 
roof  so  as  to  leave  a  level  pocket 
behind,  an  artificial  slope,  called  a 
saddle  or  cricket,  is  used.  (Fig.  195.) 

187.    Rise  and  Run.     The  slope  of  a  roof,  while  sometimes 
la'(run) — ^  stated   by   giving   the  angle  in  de- 

grees, is  more  often  expressed  in 
terms  of  rise  and  run,  as  shown  in 
Fig.  196.  The  roof  shown  in  the 
figure  would  be  said  to  have  a  rise 
of  5''  in  a  run  of  12". 
When  the  relation  of  rise  and  run  are  known,  many  relations 
of  height  can  be  deter- 
mined from  the  plan 
without  recourse  to  the 
elevation.  Relations  of 
distance  can  be  deter- 
mined on  elevation  with- 
out using  the  plan,  or  a 
plan  and  elevation  can  be 
developed  without  draw- 
ing a  roof  plan.  Thus, 
in  Fig.  197,  the  inclina- 
tion of  the  roof  being  45°, 
the  eave  sp  will  intersect 
the  roof  B,  at  the  point  p, 
on  the  plan,  which  is  just 
as  far  back  of  the  eave  inn, 
on  the  plan,  as  s  is  above 
mn  on  the  elevation. 


9 


J 


-n" 


Fig.  197 


216 


DESCRIPTIVE   GEOMETRY 


[X,  §  188 


188.    To  Determine  the  Point  at  which  a  Guy  Wire  Meets 
a  Roof.     In  Fig.  198  are  shown  the  plan  and  elevation  of  a 


Fig.  198 


building  with  a  smokestack.  Three  guy  wires  are  to  be  fastened 
to  the  stack  at  the  height  aa.  These  wires  shall  make  angles 
of  30°  with  the  ground,  and  shall  lie  in  the   vertical  planes 


X,  §  188]        APPLICATIONS  OF   CHAPTER  VI  217 

indicated  on  plan  by  OA,  OB,  and  OC.  It  is  required  to 
determine  where,  if  at  all,  these  wires  will  meet  the  roof. 

It  is  necessary  first  to  determine  the  projections  of  the  wires. 
One  wire  lies  in  the  plane  OB.  Its  H  projection,  rh^,  coincides 
with  OB,  where  r  is  chosen  as  the  point  where  the  wire  meets 
the  stack,  and  s  is  amj  point  on  the  wire.  Revolve  the  wire 
till  it  is  parallel  to  V.  Its  plan  is  now  r^s'\  Its  elevation 
passes  through  r''  and  makes  an  angle  of  30°  (the  given  angle) 
with  the  horizontal  axis.  Thus  s'  is  located.  If  now  the  wire 
is  swung  back  into  its  proper  position,  5'^'  moves  to  5^',  and 
s'^  moves  to  s^,  giving  r''s^'  as  the  true  elevation  of  the  wire. 
This  operation  is  an  application  of  §  24. 

To  find  where  the  wire  pierces  the  roof,  we  apply  §  78.  We 
require  the  trace  of  the  wire  on  the  roof.  The  auxiliary  plane 
is  chosen  through  the  wire  and  perpendicular  to  H.  It  cuts  the 
building  in  the  broken  line  t-z.  The  point  F,  where  this  Hne 
crosses  the  wire,  is  the  point  in  which  the  wire  meets  the  roof. 

The  line  tuv  is  determined  by  starting  with  the  plan.  The 
line  cb  pierces  the  auxiliary  plane  05  at  a  point  directly  above 
u^.  By  projecting  upward  from  u^,  the  V  projection  of  w  may 
be  found.  The  intersection  of  the  auxiliary  plane  with  the 
stack  is  a  vertical  line,  hence  t  lies  du-ectly  below  r  and  on  the 
intersection  between  the  stack  and  the  roof.  The  point  v  is 
not  so  easily  located  on  the  elevation,  since  it  lies  in  a  line  parallel 
to  P,  but  it  may  be  found  without  using  a  profile  projection. 
Draw  the  line  vh'\  parallel  to  the  cave  line.  This  line  is  the 
H  projection  of  a  line  in  the  roof  and  parallel  to  H.  The 
point  v'  is  located  on  elevation  by  projecting  up  from  the 
plan.     Lastly,  r^  is  found  on  a  horizontal  line  through  v''\ 

The  remaining  points  on  t-z  are  found  in  a  similar  manner. 
The  other  two  wires,  OA  and  OC,  are  not  shown  in  the  figure. 

[Note.  A  few  words  with  regard  to  the  construction  of  this  plan 
and  elevation  will  assist  in  the  apphcation  of  this  problem  to  the  one 
given  on  sheet  LXXX  on  p.  299.     All  roofs  are  of  equal  slope,   hence 


218  DESCRIPTIVE   GEOMETRY  [X,  §  188 

all  the  angles,  d,  are  45°,  and  the  line  V'cH^  is  a  straight  line  (§  185). 
The  plan  is  easily  completed,  excepting  for  the  dormers.  Leaving 
this  point,  and  passing  to  the  elevation,  the  height  of  the  ridge  ecj 
can  be  determined  from  the  known  span  and  slope.  In  this  case,  the 
slope  being  45°,  the  height  g  is  equal  to  the  distance  h.  Let  the  front 
walls  of  the  dormers  be  in  the  same  planes  with  the  walls  below.  Then 
i  =  j ',  and  if  k  is  determined  at  will,  m  must  be  made  equal  to  it. 
Thus  the  plan  and  elevation  of  one  dormer  can  be  finished.  Next 
lay  out  the  plan  of  the  dormer  which  is  seen  partly  from  the  side,  mak- 
ing it  like  the  one  just  completed.  The  elevation  of  this  one  can  be 
determined  by  projecting  corresponding  points  up  from  the  plan,  and 
across  from  the  elevation  of  the  other  dormer.  One  point  which  fre- 
quently gives  trouble  is  n.  This  point  lies  on  the  corner,  at  the  same 
height  as  p,  which  has  already  been  determined.] 

189.  Framing  for  a  Hipped  Roof.  In  framing  a  timber  roof, 
it  is  desirable  to  cut  the  timbers  so  that  they  will  fit  in  their  re- 
spective places,  before  starting  to  raise  and  assemble  them  on 
the  roof.  In  order  to  do  this,  the  various  angles  for  the  cuts 
(see  Fig.  200  a)  must  be  worked  out  from  the  plans.  The 
present  problem  takes  the  case  of  a  simple  hipped  roof  and  stand- 
ing gutter,  and  works  out  the  various  angles,  etc.,  which  must 
be  known  in  order  to  get  out  the  various  parts. 

The  roof,  as  shown  in  Fig.  199,  consists  of  three  planes,  meeting 
in  the  lines  ab,  ac,  and  ad.  It  will  be  assumed  that  the  timbering 
conforms  exactly  to  these  planes,  on  top,  and  to  similar  parallel 
planes  below.  This  will  mean  that  the  hip  and  ridge  timbers 
will  have  the  general  form  shown  in  Fig.  200  b,  though  in  an 
actual  construction  they  probably  would  be  square  edged.  In 
this  latter  case,  the  ridge  and  hips  are  slightly  blunted  in  the 
timbering,  but  the  boarding  is  brought  to  a  real  line  of  inter- 
section. 

Plax,  Elevation,  and  Section.  The  layout  of  the  plan  is 
quite  simple.  The  detail  of  the  framing,  as  shown  within  the 
circle  .1  in  Fig.  199,  should  be  carefully  noted.  The  front 
left-hand  corner  of  the  roof  is  shown  covered  with  boarding,  and 
fitted  with  a  standing  gutter.     The  section  shows  the  depth 


X,  §  1S9]        APPLICATIOXS  OF  CHAPTER  VI  219 


z'\n^y 


Fig.   199 


220 


DESCRIPTIVE   GEOMETRY 


[X,  §  189 


of  the  timbers  and  the  manner  of  fastening  them  to  the  walls, 
by  means  of  a  plate.  In  drawing  the  elevation,  it  should  be 
borne  in  mind  that  the  line  a'd'-'  is  the  V  projection  of  the  plane 
hade,  and  that  since  the  line  fg  does  not  lie  in  this  plane,  but  in 
acd,  its  F  projection, /''<7',  will  not  coincide  with  aUh'.  Similarly, 
the  lower  edges  of  the  hip,  directly  below  ad  and  Jg,  will  show 
as  two  lines,  of  which  one  is  dotted. 

In  determining  the  depth  of  the  hip  rafter,  it  should  be 
remembered  that  its  bottom  and  top  edges  lie  in  the  same 
planes  with   those  of  the   common   and   jack  rafters.     Figure 

7 


4.  Down  cut. 

5.  heel  cut. 
?.    Side  cut. 


Fig.  200 


200  c  shows  this  fact.  It  will  be  noted  that  the  line  yz  is  common 
to  both  rafters,  and  that  it  is  vertical.  Now  since  the  hip  is  set 
at  a  flatter  pitch  than  is  the  jack,  and  since  a  vertical  line  on 
either  is  of  the  same  length,  it  follows  that  the  hip  rafter  must 
be  cut  from  a  wider  plank  than  the  jack.  However,  in  the 
elevation,  this  extra  width  is  not  apparent,  and  the  hip  appears 
to  be  of  the  same  width  as  the  common  rafter  shown  in  the 
section. 

It  is  worth  while  for  several  purposes,  in  making  the  small 
scale  plan  and  elevation  indicated  in  Fig.  199,  to  study  the 
parts  within  the  circles  A  and  B  (Fig.  199)  at  a  larger  scale, 
as  shown  in  Fi^.  201,  since  the  detaih  of  these  portions  cannot 
be  drawn  accurately  on  a  small  scale. 


X,  §  189]        APPLICATIONS  OF   CHAPTER  VI 


221 


Let  us  try  to  find  : 
This  angle  is  sliown 


This  is  also  shown  in 


Cuts  and  Angles.     (See  Fig.  200  a.) 

1.  The  down  cut  for  a  common  rafter. 
in  its  true  size  on  the  section. 

2.  The  heel  cut  for  a  common  rafter. 
true  size  on  the  section. 

3.  The  side  cut  for  a  jack  rafter.  This  angle  is  shown  on  plan 
as  hNc^m^.  This  angle  does  not  show^  in  its  true  size.  Let  it 
be  rotated  on  hm  as  an  axis,  until  it  is  parallel  to  H.  During 
the  rotation,  k''  moves  to  //'"  and  k^  moves  to  k'^.  The  true 
size  of  the  angle  is  then  shown  at  h^k'hn^. 

4  and  5.  The  down  cut  and  the  heel  cut  of  the  hijp  rafter.  Draw 
an  elevation  of  the  hip  on  a   V  plane  chosen  parallel  to  the 


rafter  (§§  27,  53).     The  angles  marked  4  and  5  are  the  true 
size  of  the  required  angles. 

6.  The  top  bevel  of  the  hip  rafter.  Assume  an  H  plane  of 
reference,  as  shown  by  HA.  Extend  the  roof  planes  to  intersect 
this  plane.  The  traces  will  be  YT',  ZZ\  and  ZZ\  as  shown. 
The  line  of  intersection  of  the  planes  is  an.  Now  proceed  to 
find  the  angle  between  the  planes,  as  in  §  101. 

7.  The  side  cut  of  the  hip  rafter.  This  is  the  angle  shown  at 
fil  in  Fig.  201.  The  true  size  may  be  found  by  revolving  the 
plane  of  the  angle,  as  shown,  till  it  is  parallel  to  H. 


222 


DESCRIPTIVE   GEOMETRY 


IX,  §  189 


H 

A 

b' 

a^                  A 

\ 

-Tf- — 

X 

\                / 
\        / 
\/ 

/\ 

■■-,  Y" 

Y     _iJj  jO, 

\ 

7 

b^ ' 

t- 

\ 

\ 

~ ~1^ 

oV       -    ' '       Y' 
\ 

\x' 

y^r- 

\2. 

\ 

d^ 

A          \ 

Fig.  20C 


8  and  9.  The  down  cut  and 
ihc  side  cut  for  gutter  boards. 
Draw  an  enlarged  plan  and  ele- 
vation as  in  Fig.  202.  Rotate 
the  sides  osvp  and  ostq,  succes- 
sively, on  OS  as  an  axis,  until 
the  required  angles  are  parallel 
to  //.  This  gives  the  angles 
marked  8  and  9  in  the  figure  as 
the  true  sizes. 

10.  The  angle  between  the  gutter 
boards,  at  the  miter.  As  shown 
in  Fig.  202,  the  face  of  one  of 
the  boards,  osvp,  is  a  plane  per- 
pendicular to  I"  and  inclined  to 
// ;  while  the  face  of  the  other 
l)oard,  opxj,  is  a  plane  parallel 
to  HA.  The  line  of  intersection 
of  these  planes  is  op. 

In  Fig.  203,  the  planes  .Y  and 
Y,  and  their  line  of  intersec- 
tion ab,  reproduce  the  condi- 
tions of  Fig.  202. 
The  angle  between 
X  and  Y  being 
found,  according 
to  the  principles 
of  §  101,  the  equal 
angle  which  exists 
between  the  gut- 
ter boards  is  de- 
termined. In  the 
figure,  this  angle 
is  numbered  10. 


X,  §  190]        APPLICATIONS   OF  CHAPTER  VI 


223 


190.    A  Sheet  Metal  Chute.     The  metal  chute  shown  in  Fig. 
204  (1),  is  fastened  to  the  I  beam  by  means  of  I  olts  that  pass 


cz. 


Fig.  204 
through  a  cHp  made  from  a  bent  plate.     Given  the  size  and 


■SECTIOK 


Fig.  205 


position  of  the  chute  and  the  beam,  it  is  required  to  determine 
the  developed  pattern  of  the  clip,  and  the  angle  of  the  bend  in  it. 


224 


DESCRIPTIVE   GEOMETRY 


[X,  §  190 


In  Fig.  206,  that  part  of  the  chute  on  which  the  chp  is  to  be 
fitted  is  redrawn  at  large  scale,  in  plan  and  elevation.  Let  it 
be  assumed  that  the  clip  is  to  be  so  designed  that  its  edges 
lie  in  the  same  vertical  planes  with  the  edges  of  the  beam 

b' 


flanges.  The  lines  eg  and  fh  are  established  on  this  basis,  the 
lengths  being  taken  at  any  reasonable  amount,  and  gh  being 
made  parallel  to  ab.  The  projections  of  the  bottom  face  of 
the  clip  are  now  established. 

The  lower  face  of  the  clip  is  to  coincide  with  the  side  of  the 


X,  §  190]        APPLICATIONS  OF  CHAPTER  VI  225 

chute  and  with  the  top  of  the  I  beam ;  hence  the  angle  of 
the  bend  will  be  determined,  if  the  angle  between  the  planes 
of  the  top  of  the  beam  and  the  side  of  the  chute  is  found.  Let 
the  top  of  the  beam  flange  be  taken  as  the  H  plane,  establishing 
HA  as  shown.  The  H  traces  of  ah  and  cd  are  at  j  and  h  re- 
spectively. These  points  determine  the  line  jm  which  is  the 
H  trace  of  the  side  of  the  chute.  The  line  hi,  a  part  of  jm,  is 
the  line  along  which  the  clip  must  be  bent. 

Now  let  ZZ^  be  drawn  to  represent  the  H  trace  of  a  cutting 
plane,  passed  perpendicular  to  jm.     Such  a  plane  will  cut  the 
side  of  the  chute  in  the  line  no,  and  the  angle  ony      ^_______^ 

will   be   the  supplement  of   the   required   angle   for      V®"^  \ 
bending   the  clip.     The   true   size   of   this   angle   is      /  ^^n;  ] 

found  by  rotating  the  triangle  nop  on  the  side  np     hi /^ 

until  o  falls  in  H,  at  o^.     (o'^o^  is  perpendicular  to    / '?  -  o  / 
ZZ^  and  is  equal  to  p^'o^\)  yig.  lu7 

'    To  find  the  development  of  the  clip,  let  the  lines 
kg  and  Ih  be  rabatted  about  jm,  into  H,  giving  e^fH^h'^g'^lc^  as  the 
development  of  the  under  side  of  the  clip.     By  rabatting  s,  in 
the  same  manner,  the  line  y^s'^'  can  be  established  to  show  the 
line  of  contact  between  the  edge,  cd,  of  the  chute,  and  the  clip. 

In  Fig.  207,  the  development  of  the  clip  is  redrawn  and  the 
bolt  holes  are  located.  The  exact  location  of  the  holes  and  the 
necessary  allowances  for  bending  are  matters  of  technical 
detail  that  need  not  be  discussed  here.  It  should  be  noted,  ^ 
however,  that  if  the  edges  of  the  clip  are  to  be  perfectly  vertical 
planes,  the  sides  kg  and  Ih  must  be  slightly  beveled  before 
bending. 


CHAPTER   XI 


APPLICATIONS   BASED    ON    CHAPTERS   VII   AND    Villi 


191.    To  Determine  the  Appearance  of  the  Grain  on  a  Turned 
Wooden  Vase.     Let  a-h,  Fig.  208,  be  the  block  from  which  the 

vase  is  to  be  turned.  Let  the 
annual  rings  be  assumed  to  take 
the  form  of  cylinders,  as  shown  on 
plan  by  the  concentric  circles,  1-8. 
Let  it  be  required  to  trace  the 
emergence  of  these  rings,  on  the 
surface  of  the  vase,  as  shown  on 
the  elevation. 

Any  horizontal  cutting  plane,  as 
A,  will  cut  the  vase  in  a  circle  as 
shown  at  m^'-x-f'-  on  plan.  On  the 
front  side  of  the  vase,  this  surface 
circle  cuts  the  cylinders  forming 
the  grain,  at  points  j,  k,  J,  and  m. 
These  points  are  determined  on 
plan  and  projected  to  the  eleva- 
tion. Other  cutting  planes  would 
give  other  points.  By  joining, 
on  the  elevation,  all  such  points 
which  result  from  the  same  grain 
cylinder,  the  appearance  of  the 
grain  can  be  established  readily. 

Where  a  part  of  the  vase  consists  of  a  ^•ertlcal  cylinder,  the 
grain  lines  will  be  vertical  straight  lines.      The  points  where 

1  Exercise  sheets  to  accompany  this  chapter  are  given  on  pages  300  304. 

226 


Fig.  208 


XI,  §  192]    APPLICATIONS   OF   CHAPTERS  VII-VIII    227 

any  given  line  crosses  the  meridian  section,  as  n  and  o,  can  be 
determined  exactly  by  noting  on  plan  the  intersection  of  the 
meridian  plane,  M,  and  the  grain  cylinder  under  consideration, 
No.  5.  The  point  of  tangency  between  the  surface  circle  A 
and  the  grain  cylinder  6,  as  seen  on  plan  at  m^,  indicates  that 
on  the  elevation  the  grain  line  just  touches  the  plane  of  the 
surface  circle. 

192.    A  Spiral  Chute.     Figure  209  shows  a  spiral  chute  made 
of  sheet  iron  for  lowering  packages.      The  vertical  side  forms 


Fig.  209 

a  portion  of  a  cylinder.  The  bottom  is  a  right  helicoid,  gen- 
erated by  an  horizontal  line  connecting  two  helixes  of  different 
diameters. 

The  slope  of  the  inner  helix  is  much  steeper  than  that  of  the 
outer  one.  Light  packages  descend  along  the  steep  inner  edge, 
while  heavier  ones,  under  the  effect  of  centrifugal  force,  work 
themselves  out  toward  the  lower  gradient  of  the  outer  edge. 

The  drawing  of  the  plan  and  elevation  involves  the  construc- 
tion of  several  helixes  (§  137).  A  developed  pattern  may  be 
constructed  for  the  vertical  side  following  the  principles  of 
§  157.  The  bottom,  being  a  warped  surface,  may  only  be  ap- 
proximated by  development  (§  161). 


228  DESCRIPTIVE   GEOMETRY  [XI,  §  193 

193.    Intersecting  Vaults. 

I.  Given  the  Cross-Section  of  two  Intersecting  Barrel 
Vaults,  to  Determine  the  Line  of  Intersection.  A  barrel 
vault  is  a  hollow  cylinder,  the  cross  section  of  which  may  be 
of  any  desired  shape.  The  most  usual  form  of  section  is  semi- 
circular, though  elliptical,  three  centered,  and  pointed  arch 
forms  are  not  uncommon. 

Figure  210  shows  two  intersecting  semicircular  barrel  vaults, 
drawn  without  thickness.  These  vaults  intersect  in  the  space 
curves  ahc  and  a'b'c'.  The  problem  of  determining  such  a 
curve,  on  plan  and  elevation,  is  merely  that  of  finding  the  line 
of  intersection  of  two  cylinders,  as  described  in  §§  149  and  152. 


Figure  211  shows  the  plan  and  two  elevations  of  intersecting 
vaults  similar  to  those  shown  in  Fig.  210.  Here  the  thickness 
is  indicated.  The  line  of  intersection  of  the  outer  surface  of 
the  shell  is  determined  first.  Cutting  planes  parallel  to  H  are 
used.  One  such  plane,  X,  is  shown  in  the  figure.  This  plane 
cuts  the  large  vault  along  two  elements  which  are  projected 
on  elevation  in  the  points  d''  and  /' .  These  elements  are  shown 
on  plan  at  d^e''  and  f^g^. 

The  same  cutting  plane,  X,  also  cuts  the  smaller,  or  trans- 
verse vault,  along  two  elements.  These  elements  are  projected 
in  the  points  h^  and  k^  on  the  end  elevation,  and  in  the  lines 
h''  and  /  and  JcH^  on  plan. 

These  two  pairs  of  elements  are  in  Ae  same  plane,  A^,  and  are 
not  parallel.  Hence  they  will  intersect.  The  points  of  inter- 
section, m,  n,  o,  and  p,  arc  on  the  surfaces  of  both  vaults,  and 


XI,  §  193]     APPLICATIONS   OF  CHAPTERS  VII-VIII    229 

hence  are  points  on  the  hne  of  intersection  of  the  vaults.  Other 
cutting  planes,  parallel  to  X,  would  give  other  points,  helping 
to  determine  the  curve  amhnc. 

The  line  of  intersection  of  the  inner  surface  of  the  vaults  is 
shown  as  a  dotted  Hne.  This  line  is  determined  in  the  same 
manner   as   above.     The   cutting   plane,    X,   locates    the   four 


Fig.  211 

elements  which  are  shown  by  the  dotted  lines.     These  elements 
in  turn  locate  the  four  points  qrst. 

The  two  curves  thus  determined  are,  in  general,  some  form 
of  space  curve.  When  the  intersecting  vaults  are  semicir- 
cular, and  their  centers  are  at  the  same  height,  the  plan  of 
the  hne  of  intersection  is  an  hyperbola.  If  the  vaults  are  of 
the  same  diameter,  and  their  centers  are  at  the  same  height, 
the  line  of  intersection  becomes  a  pair  of  ellipses,  and  the  plan 
becomes  a  pair  of  straight  lines. 


230 


DESCRIPTIVE   GEOMETRY 


[XI,  §  193 


II.  Given  one  Vault  and  its  Line  of  Intersection  with 
Another  Vault,  to  Determine  the  Cross-section  of  the 
Second  Vault.  In  Fig.  212,  the  intersecting  vaults  are  shown 
as  without  thickness.  Let  the  plan  and  the  front  elevation 
be  given  in  full,  and  let  it  be  required  to  establish  the  cross 
section  as  shown  on  the  side  elevation. 

The  cutting  plane,  Z,  determines  the  element  mn  on  the  larger 
vault  and  hf  and  dk  on  the  smaller  vault.     These  lines,  pro- 


PUN 


Fig.  212 


jected  to  the  side  elevation,  locate  the  points  h^  and  d^  which 
are  two  of  the  points  required  to  establish  the  curve  a^h^c^dPe'^. 

194.  Mitering  of  Mouldings.  A  moulding  is  a  combination 
of  cylindrical  and  plane  surfaces.  Its  right  section  or  profile 
is  therefore  a  combination  of  straight  and  curved  lines. 

Plaster  and  cement  mouldings  are  formed  by  a  knife  which 
is  slipped  along  on  straight  guides  and  in  contact  with  the 
plastic  material.  In  running  a  wooden  moulding,  the  knife 
revolves  while  the  material  is  shoved  past  it,  on  a  straight  guide. 


XI,  §  194]     APPLICATIONS   OF  CHAPTERS   VH-VIII    231 


If  a  moulding  is  run  on  a  curved  surface,  it  becomes  a  combina- 
tion of  cylindrical  and  double-curved  surfaces.  Since  such  cases 
are  relatively  rare,  the  following  discussion  is  limited  to  mouldings 
having  straight 
elements.  The 
principal  prob- 
lems on  intersec- 
tions of  mould- 
ings depend  on 
the  ideas  stated 
in  §§  138-152. 

If  a  piece  of  moulding  is  cut  by  planes  making  equal  angles 
to  right  and  left  with  its  elements,  as  shown  in  Fig.  213,  the 
sections  a  and  b  are  equal.  Therefore  the  end  pieces,  A  and  B, 
placed  together  as  shown  at  (2),  will  match  exactly,  forming  a 
miter.     In  Fig.  213  (2),  the  elements  of  the  moulding  are  turned 


Fig.  213 


through  90°.  This  is  the  usual  case,  but  any  desired  angle  may 
be  secured  at  the  miter  by  varying  the  angle  between  the 
cutting  planes  and  elements,  as  shown  in  Fig.  213  (3). 

If  the  cutting  plane  is  inclined,  not  only  toward  the  elements, 
but  also  toward  the  plane  of  the  base,  as  shown  in  Fig.  214 
(1),  the  mouldings  will  miter  as  shown  in  214  (2). 

Besides  mitering,  mouldings  may  also  be  brought  to  an 
intersection  by  coping,  as  shown  in  Fig.  214  (3).  The  appear- 
ance as  seen  from  the  finished  side  is  the  same  in  either  case. 


232 


DESCRIPTIVE   GEOMETRY 


[XI.  §  194 


Fig.  215 


Mouldings  that  have  different  profiles  may  be  made  to  meet 
in  various  ways  under  varying  conditions.     The  cases  described 

in  §§  195-197  are 
selected  as  being 
typical. 

195.  To  Find  the 
Line  of  Intersection 
of  Two  Mouldings  of 
Different  Profiles. 
In  Fig.  215,  let  the 
plan  and  the  profiles 
1  and  2  be  given, 
and  let  it  be  required 
to  find  the  plan  and 
the  elevation  of  the 
line  of  intersection. 
A  series  of  horizontal  cutting  planes,  A,  By  C,  is  used  to  estab- 
lish a  corresponding  series  of  elements  on  plan  and  on  elevation. 

Thus  the  plane  D  determines 
the  element  m  on  moulding  2, 
and  element  m'  on  moulding  1. 
On  plan,  these  two  elements  are 
seen  to  intersect  at  m"^.  By 
projecting  up  from  m"^  to  the 
plane  D,  on  the  elevation,  m"^ 
is  determined.  One  point  on  the 
line  of  intersection  is  now  estab- 
lished. The  principle  involved  is 
identical  with  that  of  §  193. 

196.  Raking  Mouldings.  Fig- 
ure 216  shows  a  small  section  of 

a  typical  classic  pediment.  Two  types  of  mitering  are  illus- 
trated. Mouldings  a  and  h  have  the  same  profile,  and  miter 
on  a  45°  plane  at  the  corner,  in  the  usual  manner.     The  joining 


XI,  §  196]    APPLICATIONS   OF   CHAPTERS  VII-VIII    233 


of  c  and  d,  however,  involves  a  different  principle.  The  level 
moulding,  c,  is  made  not  only  to  turn  the  corner  but  also  to 
change  its  inclination  to  H  at  the  same  time.  The  moulding  c  is 
called  the  level  moulding,  while  d  is  called  the  rake  moulding. 

The  intersection  of  a  level  moulding  with  a  rake  moulding  is 
usually  managed  in  one  of  two  ways.  These  ways  are  illustrated 
in  Fig.  217.  At  (1),  two  blocks,  m  and  n,  are  shown  mitering 
in  the  usual  manner.  At  (2),  the  same  blocks  have  been 
tipped,  rotating  on  cd,  so  that  n  is  in  the  position  of  a  raking 
moulding.  The  blocks  have  the  same  section  as  before.  They 
still  meet  perfectly  on  the  mitering  plane,  but  the  mitering 
plane  is  no  longer  at  an  angle  of  45°  to  T^,  and  90°  to  H.  More- 
over, the  horizontal  and  vertical  faces  of  m  are  thrown  into  slop- 
ing positions.     The  effect  is  unpleasant,  since  the  level  moulding 


has  a  greater  projection  than  the  rake  moulding,  and  the  miter- 
ing plane  does  not  bisect  the  angle  of  the  wall. 

A  more  difficult  but  more  effective  method  is  shown  at  (3). 
Here  the  block  m  remains  in  the  same  position  as  in  (1),  and  the 
mitering  plane  is  passed  in  the  usual  manner.  The  cross  section 
of  the  block  ?i'  is  determined  so  that  it  will  meet  accurately  with 
m,  while  following  the  required  rake.  In  this  case,  vertical  sur- 
faces on  the  level  moulding  intersect  perfectly  with  vertical  sur- 
faces on  the  rake,  though  the  widths  are  different.  Horizontal 
surfaces,  however,  do  not  intersect  so  well,  as  will  be  shown  later. ^ 

1  Classic  examples  involving  rake  moulding  are  usually  worked  out  on 
the  principle  of  Fig.  217  (3).  It  is  interesting  to  note,  in  these  examples 
(see  Fig.  216),  how  carefully  the  intersection  of  all  horizontal  surfaces  of 
the  level  moulding  with  corresponding  surfaces  on  the  rake  moulding  has 
been  avoided. 


234 


DESCRIPTIVE   GEO:\IETRY 


[XI,  §  196 


The  method  of  determining  the  profile  required  for  a  rake 
moulding  is  quite  simple  when  the  angle  of  the  wall,  ejg,  Fig. 
216,  is  90°.  When  the  angle  of  the  wall  is  not  90°,  the  solution 
is  more  complex.  Both  cases  are  given  in  the  following 
article. 

197.    To  Determine  the  Profile  of  a  Rake  Moulding. 
I.   When  the  Angle  of  the  Wall  is  90°.     In  Fig.  218, 
let  the  profile  of  the  level  moulding  be  given  as  ahfg,  and  let  the 

pitch  of   the  rake  be  given 
as  (9. 

Let  the  angle  of  the  wall, 
fal*m^,  and  the  trace  oi  the 
mitering  plane,  AA^,  be 
drawn.  Next  draw  the  plan 
of  the  element  aj  and  as 
many  others  as  are  deemed 
necessary.  These  elements 
intersect  in  the  mitering 
plane  at  the  points  a,  b,  etc. 
The  plan  of  the  rake  mould- 
ing can  now  be  drawn  from 
a^,  h^,  etc.,  and  parallel  to 
a^m^.  The  elevation  of  the 
element  am  is  drawn  next, 
at  the  required  pitch,  the 
other  elements  being  parallel  to  it.  Thus  the  plan  and  elc\n- 
tion  of  the  rake  moulding  are  determined.  It  remains  to  deter- 
mine the  right  section. 

Pass  the  plane  T  perpendicular  to  the  elements  of  the  rake 
moulding.  This  plane  cuts  the  moulding  in  the  required  section, 
the  V  projection  of  which  is  Jq.  Rabatte  the  points  l-q  into 
H,  as  shown  in  the  figure.  The  points  thus  found  determine 
the  required  curve. 

The  right  angle  at/,  on  the  level  moulding,  corresponds  to  an 


XI,  §  198]    APPLICATIONS  OF  CHAPTERS  VII-VIII     235 

obtuse  angle  on  the  rake  moulding.     In  large  surfaces   this 
would  probably  be  unpleasant.     (See  footnote,  page  233.) 

II.  When  the  Angle  of  the  Wall  is  not  90°.  The  prin- 
ciple involved  in  this  case  is  the  same  as  for  case  I.  The  method 
differs  only  in  detail.  The  plan  and  elevation  of  the  rake 
moulding  are  established  in  the  same  manner  as  in  the  former 
case.  Next  a  plane  is  passed  perpendicular  to  the  elements  of 
the  rake  moulding  (§  81).  This  plane  will  cut  the  required 
right  section  from  the  moulding.  Next  find  where  each  element 
of  the  moulding  pierces  the  cutting  plane  (§78).  By  connecting 
the  points  thus  found,  curves  are  determined  w^hich  are  the 
projections  of  the  required  right  section.  The  true  shape  of 
the  right  section  can  then  be  found  by  rabattement  (§  91). 


Fig.  219 


198.  Domes  and  Pendentives.  Usually  a  dome  is  some  form 
of  double-curved  surface,  spheres  and  spheroids  being  the  more 
usual  forms.  The  simplest  case  is  that  of  a  hemisphere  sup- 
ported on  a  cylindrical  case,  as  in  the  Pantheon  at  Rome.  It 
is  often  necessary,  however,  to  support  a  dome  on  a  base  which 
is  a  square,  octagonal,  or  other  form  of  prism.  When  this 
happens,  some  special  means  must  be  employed  to  make  the 
transition  between  the  dome  and  its  supporting  walls  or  piers. 

In  Fig.  219,  one  of  the  simplest  cases  is  shown.  Here  the 
diameter  of  the  dome  is  made  equal  to  the  diagonal  of  the 
square  base,  and  the  planes  of  the  base  are  allowed  to  pass 
upward,  cutting  the  dome  in  the  circular  arcs,  ab,  he,  etc.     If 


236 


DESCRIPTIVE  GEOMETRY 


[XI,  §  198 


it  is  desired  to  reduce  the  supporting  walls  to  four  piers,  it  can 
be  done  by  substituting,  in  place  of  the  walls,  four  arches, 
sprung  between  the  piers,  as  shown  in  Fig.  220.  So  far  as  the 
dome  is  concerned,  the  shape  is  the  same  whether  it  is  carried 
on  walls  or  on  piers  and  arches.  An  example  of  this  type, 
slightly  modified,  is  to  be  found  in  the  Baptistery  at  Ravenna. 

Another  method  of  supporting  a  dome  over  a  square  plan 
is  shown  in  Fig.  221,     Here  the  diameters  of  the  dome  and  of 


•HALf  PIAN      LOOKING   UP- 
Fig.  220 

the  square  base  are  equal.  This  construction  leaves  the  spaces 
ahc,  etc.,  which  are  not  covered  by  the  dome,  and  the  dome  is 
not  supported  by  the  base  along  the  lines  ac,  etc.  The  necessary 
coverings  and  supports  are  furnished  by  filling  in  these  spaces 
in  one  of  several  ways.  These  filling  and  supporting  parts  are 
called  the  pendentives  of  the  dome. 

In  Fig.  221  the  pendentives  are  parts  of  a  sphere  whose  di- 
ameter is  equal  to  the  diagonal  of  the  plan.  (See  dbe,  Fig. 
219.)  The  well-known  dome  of  Santa  Sophia  in  Constantinople 
is  supported  on  pendentives  of  this  type. 

Sometimes  the  support  of  a  dome  is  in  the  form  of  an  octagon, 


XI,  §  199]    APPLICATIONS   OF  CHAPTERS  VII-VHI    237 


carried  to  a  square  plan  by  means  of  plane,   cylindrical  or 
conical  pendentives,  or  merely  by  corbeling. 

The  jointing  planes  of  stonework  are  kept  normal  to  the 
exposed  faces,  in  so  far  as  possible,  in  order  to  avoid  thin 
edges  on  individual  stones.  It  naturally  follows  that  the 
horizontal  joints  in  a  spherical 
dome  are  formed  along  the  sur- 
faces of  cones  whose  apexes  are  at 
the  center  of  the  sphere,  while  the 
vertical  joints  follow  meridian  sec- 
tions, as  shown  in  Fig.  221. 

199.  The  Decorative  Treatment 
of  Curved  Surfaces.  Decoration 
may  be  classed  as  geometrical  when 
it  is  composed  of  regular  lines, 
repeated  in  a  set  manner,  as  in 
diapers,  frets,  over  all  patterns, 
etc.,  or  as  free,  when  it  represents 
figures  or  other  natural  forms, 
without  repeats. 

Geometrical  decoration  usually 
follows  some  element  or  other  char- 
acteristic line  of  the  surface  to  be 
decorated ;  hence  it  can  be  accu- 
rately shown  at  small  scale  in  the 

usual  manner.  The  more  naturalistic  the  forms  used,  however, 
the  more  difficult  is  any  adequate  representation  at  small  scale. 
The  cartoons  for  free  ornament  must  then  be  made  at  full  size, 
by  skilled  artists.  Hence  there  is  a  tendency  to  have  such 
work  done  in  studios,  on  flexible  backings  of  paper  or 
canvas,  these  being  later  set  in  place  complete.  Of  course, 
this  manner  of  execution  is  only  possible  when  the  surfaces 
to  be  treated  are  developable.     (See  also  §  144.) 

In  designing  the  decoration  for  a  developable  surface,  the 


Looking  down— ^Looking  up- 
Fig.  22  i 


238 


DESCRIPTIVE   GEOMETRY 


[XI,  §  199 


development  is  drawn  first,  the  pattern  being  worked  out  on  the 
flat,  so  that  they  fit  it  perfectly  and  repeat  or  work  out  all 
around  as  may  be  desired. 

For  double-curved  or  warped  surfaces,  the  decoration,  if 
geometrical,  may  be  designed  by  using  a  plan  and  elevation 
or  section,  though  it  is  somewhat  difficult.     But  for  naturalistic 


Plan  -  looking  up. 


Development  of  one  half 
larger   voult. 

Fig.  222 


representations,  it  is  necessary  to  w^ork  on  a  three  dimensional 
model,  either  at  scale  or  in  full  size,  unless  a  satisfactory  ap- 
proximate development  (§§  157  and  161)  can  be  made. 

200.  To  Show  a  Geometric  Pattern  on  a  Barrel  Vault.  Let 
two  intersecting  barrel  vaults  be  given,  as  shown  in  Fig.  222,  and 
let  their  line  of  intersection,  ahc,  be  found  by  §  193.  Next 
let  the  development  of  both  be  drawn,  as  explained  in  §  157. 
It  will  be  noticed  that  the  outlines  of  the  penetration,  a'h'c'. 


XI,  §  200]     APPLICATIONS   OF   CHAPTERS   VH-VHI     239 

and  a"h"c" ,  are  not  of  the  same  form  in  development.     Ob- 
viously the  two  lines  must  be  of  equal  length. 

In  selecting  a  unit  for  the  repeated  pattern,  it  should  be  borne 
in  mind  that  the  size  chosen  must  be  common  divisor  of  the 
circumferences  of  the  two  vaults.  In  the  figure,  the  diagonal 
of  the  square  forming  the  pattern  is  contained  six  times  in  the 
smaller  and  eight  times  in  the  larger  vault. 

Now  let  the  desired  pattern  be  drawn  on  the  developments, 
and  let  guide  lines,  as  x'y'  and  y'ci ,  be  drawn  to  assist  in 
transferring  the  pattern  to  the  plan.  These  guide  lines,  being 
elements  of  the  surface,  can  be  located  on  the  plan,  as  at  xy 
and  'pq,  and  their  intersections  can  then  be  used  as  guide  points 
for  drawing  the  pattern.  In  the  illustration,  the  main  lines  of 
the  pattern,  being  diagonal  on  the  development,  appear  as 
helixes  on  the  plan. 

It  will  be  noticed  that  the  pattern  chosen  does  not  work  out 
well  along  the  penetration  line.  This  is  due  to  the  fact  that 
the  curves,  a'h'c',  and  a"h"c" ,  Fig.  222,  do  not  cut  the  patterns 
at  corresponding  points.  ]\Ioreover,  it  is  improbable  that  any 
set  pattern  could  be  devised  that  would  connect  properly  along 
the  given  line  of  intersection.  This  difficulty  can  be  minimized 
by  introducing  a  plain  band  along  the  penetration,  as  at  ah  in 
the  illustration.  The  particular  pattern  used  in  the  illustration 
is  the  same  as  that  in  the  Baths  of  Caracalla  and  in  the 
Pennsylvania  Terminal  in  New  York. 

It  would  be  possible  to  determine  the  cross  section  of  the 
smaller  vault  in  such  a  manner  that  a  pattern  could  be  worked 
out  on  it  which  would  be  similar  to  that  on  the  larger  vault, 
and  the  two  patterns  would  intersect  perfectly  on  the  pene- 
tration. Such  an  attempt,  applied  to  the  case  shown  in  Fig. 
222,  gives  for  the  outline  of  the  smaller  vault  a  pointed  curve, 
awkward  in  line  and  quite  impossible  to  use.  The  shape  of 
the  vaults,  being  the  more  important,  will  ordinarily  be  de- 
termined without  reference  to  the  ornamentation. 


240 


DESCRIPTIVE   GEOMETRY 


[XI,  §  201 


201.  Warped  Surfaces.  In  abstract  problems,  warped  sur- 
faces can  be  described  best  by  means  of  their  directing  lines  or 
surfaces,  in  one  of  the  combinations  given  in  §  161.  But  in 
practice  it  is  often  convenient  to  substitute  a  time  element  in 
place  of  one  of  the  space  elements  required  to  define  the  surface, 
by  specifying  the  rate  of  motion  of  the  generatrix. 

Thus  Fig.  223  shows  the  plan  and  elevation  of  a  recessed 
doorway  with  a  segmental  head.     The  curves  ad  and  be  can 

be  used  as  directrices  for  a 
warped  surface  covering  the 
recess.  The  lines  ab  and  cd  evi- 
dently must  be  elements  of  such 
a  surface.  The  other  elements 
may  be  determined  by  setting 
the  condition  that  the  rate  of 
motion  of  the  generatrix,  be- 
tween the  positions  ab  and  cd, 
shall  be  uniform.  This  results 
in  elements  spaced  at  equal  in- 
tervals along  each  curve,  i.e. 
ae  =  cf=fg,  and  bh  =  hi=ij,  etc. 
If  such  a  surface  is  cut  by  any 
plane,  as  Z,  which  cuts  all  the  elements,  the  line  of  intersection, 
ac,  may  be  used  as  a  directrix.  This  new  directrix,  taken  with 
the  other  two,  will  define  the  surface  quite  as  well  as  the  rate 
of  motion  described  above.  However,  the  method  using  the 
rate  of  motion  is  usually  more  convenient. 

The  problem  which  arises  most  frequently  in  connection 
with  warped  surfaces  is  to  fix  the  character  of  the  intervening 
surface  when  two  directrices  are  given  as  limits.  One  such 
case  has  already  been  shown  in  Fig.  223.  Another  is  shown  in 
Fig.  224.  Here  a  semicircle  and  a  semi-ellipse  in  parallel  planes 
constitute  two  directrices  for  the  surface.  The  case  is  such 
as  might  arise  in  designing  a  splayed  archway,  when  the  height 


Fig.  223 


XI,  §  201]     APPLICATIONS   OF  CHAPTERS  VII-VIII     241 


of  the  opening  must  be  the  same  inside  and  out,  both  at  the 
crown  and  at  the  springing  hne. 

Four  methods  of  generation  are  shown  and  others  are  pos- 
sible, of  course.  In  order  to  assist  in  a  clear  conception  of  the 
difference  between  the  surfaces  shown,  an  auxiliary  section  has 
been  drawn,  inside  the  wall.     Case  III  would  be  impossible  if 


Direcirices  •.            A-    ,,,       '       •  ■  -                   J'.  ,; 

abc  ,cief          /Vr'    r«\y_     abc  ,  def  ghi  ,  kim         _"/     v-^^^'.;,     ghi  ,  kirr, 

and  op."         /'!','.' i    " '•~.vV'    qnd  -ox  and    H.        /u' >      ."A   and  uniform 

-''//','    1   \i.,.    ,      '  •      I      ,^^""  ,      ,  i\«^  rareoTmo^- 
;       :/''/;;  :j    ;     ';>.       Auxiliar:^  verticil     ;    >''//;     ',  '•  VV.  ion  on  each. 

,  /,,'/.  '  I        ,      I  ,'•  .  '' cuTTinq  plane,  ^r^    i  ;'.  /  -    '  f          '  >  !>•  i 

..^    X',';  '      ,         \ -      Y    '  — 1-tr    /'/ '       I     !.t  _ *^^^ — 


':h^ 


the  crowns  of  the  two  curves  were  at  different  levels,  but  the 
other  cases  are  possible  for  any  combination  of  arches. 

It  is  not  at  all  unusual  to  run  one  warped  surface  into  another, 
the  two  having  one  common  element  but  quite  different  direc- 
trices. One  such  case  is  shown  in  the  covering  over  the  recessed 
gateway  in  Fig.  178.  When  this  is  done  care  should  be  taken 
to  insure  that  the  surfaces  join  smoothly.  This  can  be  done  by 
passing  an  auxihary  plane  through  the  part  in  question  to  see 
if  it  gives  a  smooth  curve  of  intersection. 


CHAPTER  XII 
SHADES   AND    SHADOWS » 

202.  Introduction.  Except  to  the  trained  eye,  the  elevations 
of  a  building  convey  no  adequate  picture  of  the  proposed 
structure,  since  the  modeling,  i.e.  the  projections  and  recessions 
from  the  main  plane  of  the  fa9ade,  is  nowhere  shown  on  a  single 
drawing. 

An  actual  building,  viewed  through  a  fog  or  in  a  very 
diffused  light,  shows  much  the  same  quality  of  flatness  as  an 
elevation  drawing.  But  in  sunlight  the  modehng  at  once 
becomes  clear  because  of  the  contrast  between  unequally  lighted 
surfaces.  In  preparing  drawings  for  exhibition,  it  is  usual  to 
imitate,  as  clearly  as  may  be,  this  effect  of  sunlight  and  shadow, 
in  order  to  assist  the  layman  to  a  correct  understanding  of  the 
work  in  hand.  Hence  the  ability  to  correctly  indicate  the 
shades  and  shadows  on  a  drawling  is  an  important  part  of  the 
draftsman's  equipment. 

This  chapter  is  not  intended  to  give  a  complete  knowledge 
of  the  casting  of  shadows.  It  is  intended  merely  to  lay  a  sound 
foundation  for  a  more  extended  course  by  emphasizing  the  fact 
that  shadow-casting  is  merely  a  special  application  of  De- 
scriptive Geometry.  For  this  reason  short  cut  methods  have 
been  avoided  and  in  many  cases  the  text  is  abridged  by  refer- 
ences to  the  basic  problems  that  have  been  explained  in  Chapters 
I  to  VIII. 

203.  Fundamental     Ideas     and    Definitions.     The     factors 

involved  in  any  pro])lem  in  shades  and  shadows  are:    (1)  a 

source  of  light;    (2)  a  body  intercepting  the  light;    and  (3)  a 

1  Exercise  Sheets  to  accompany  this  chapter  are  given  on  pp.  304-309. 

242 


XII,  §  203] 


SHADES  AND  SHADOWS 


243 


receiving  surface  on  which  the  shadow  falls.  These  factors 
are  illustrated  in  Fig.  225.  From  the  source  of  light  L,  issue 
rays  that  illuminate  the  sphere  and  the  plane  P,  beyond  it. 
Those  rays  of  light  which  fall  on  the  sphere  are  intercepted  by 
it  and  do  not  fall  on  P.  All  such  rays  are  included  within  a 
cone  tangent  to  the  sphere,  whose  apex  is  at  L.     The  frustum 


Fig.  225 

of  this  cone  which  lies  between  the  sphere  and  P,  is  a  dark 
space  called  the  umbra. 

The  line  of  tangency  between  the  cone  and  the  sphere  is 
called  the  shade  line,  or  separatrix,  and  all  that  part  of  the  sphere 
which  lies  within  the  umbra  is  said  to  be  in  shade.  The  inter- 
section of  the  receiving  surface  with  the  umbra  defines  the 
shadow.  It  is  well  to  note  carefully  the  following  distinction : 
a  surface  in  shade  is  a  part  of  the  intersecting  body  which 
is  turned  away  from  the  source  of  light,  while  a  surface  in 


244 


DESCRIPTIVE   GEOMETRY 


[XII,  §  203 


shadow  is  a  part  of  the  receiving  surface  from  which  the 
Hght  is  excluded.  (See  Fig.  225.)  In  some  cases  (see  Fig. 
230),  a  complex  body  may  be  partly  in  shade  and  partly  in 
shadow. 

In  this  book,  we  cannot  deal  extensively  with  the  problem  of 
intensity  of  illumination.  Suffice  it  to  say  that  the  intensity 
of  illumination  will  be  greatest  on  those  surfaces  which  are 
normal  to  the  rays  of  light,  and  zero  on  those  surfaces  which 
are  tangent  to  the  light  rays.  All  conceivable  intermediate 
cases  exist.  Shaded  surfaces  are  usually  illuminated  to  some 
extent  by  reflected  light. 

204.  Source  of  Light.  In  such  a  case  as  that  shown  in  Fig. 
225,  it  is  evident  that  if  the  source  of  light  is  moved,  the  shadow 


Fig.  226 


will  change  its  form  or  its  position  or  both.  If  the  source  is 
moved  to  an  infinite  distance,  the  tangent  cone  becomes  a 
tangent  cylinder,  and  the  rays  of  light  are  all  parallel.  This 
is  what  happens,  in  effect,  when  the  sun  is  taken  as  the  source 
of  light. 

In  order  to  simplify  the  problem  of  casting  shadows,  as  well 
as  that  of  interpreting  shadows,  it  is  usual  to  assume  the  sun, 
in  a  fixed  position,  as  the  source  of  light.  The  conventional 
position  selected  is  above,  in  front  of  and  to  the  left  of  the  object 
which  is  being  drawn,  the  precise  position  being  such  that  any 
ray  of  light  follows  the  diagonal  of  a  cube,  whose  faces  are 
parallel  to  the  planes  of  projection,  as  shown  in  Fig.  226'.     It 


XII,  §  205]  SHADES  AND   SHADOWS  245 

will  be  noted  that  each  of  the  projections  ^  of  such  a  ray  makes 
an  angle  of  45°  with  HA.  Thereby  two  things  are  accom- 
plished; first,  in  drawing  any  ray  of  light  either  on  plan  or 
elevation,  the  45°  triangle  may  be  used;  and  second,  the 
shadow  cast  on  any  object  by  a  projecting  part  will  show,  by 
its  width,  the  exact  amount  of  the  projection.  Thus  in  Fig.  230, 
the  width  of  the  shadow,  1,  is  equal  to  that  of  the  projection,  2. 

In  special  cases,  for  special  reasons,  the  position  of  the  sun 
may  be  varied  at  will  from  the  above,  or  any  nearer  source  of 
light  may  be  chosen.  In  the  following  problems  the  source  of 
light  is  varied  in  order  to  avoid  the  tendency  toward  the  use 
of  rules  and  formulas. 

205.  Fundamental  Analysis.  The  rays  of  light  tangent  to 
any  object  form  an  envelope  (§  142)  for  that  object.  The  line 
of  intersection  between  this  envelope  and  any  surface  receiving 
light,  is  the  outline  of  the  shadow. 

In  the  case  of  sun  shadows,  the  envelope  is  a  cylinder,  a  prism, 
or  some  combination  of  cylinders  and  prisms.  For  near-by 
sources  of  light  the  envelope  is  a  cone,  a  pyramid,  or  some  com- 
bination of  cones  and  pyramids. 

It  follows  that  the  determination  of  the  shade  and  shadow  for 
any  object  is  a  problem  in  Descriptive  Geometry.  Xo  special  rules 
or  formulas  are  needed  in  any  case.  However,  in  many  special 
cases,  the  solutions  become  complex  and  tedious.  For  such 
cases  special  short  cut  methods  may  be  established.  These 
methods,  however,  are  properly  the  subject  for  a  separate  course 
in  shades  and  shadows.  For  the  problems  which  follow,  it  is 
sufficient  to  determine  the  umbra  by  lines,  planes,  cylinders, 
cones,  etc.,  that  pass  through  the  source  of  light  tangent  to  the 
body.  The  intersection  between  this  umbra  and  the  receiving 
surface  is  found  by  the  usual  methods  for  finding  intersections 
of  surfaces. 

1  Let  the  student  determine,  both  analyticallj'  and  graphically,  the  true 
angle  that  such  a  ray  makes  with  H  or  T'. 


246 


DESCRIPTIVE  GEOMETRY 


[XII,  §  206    ]  / 


206.  Shadow  of  a  Point. 

I.  When  the  Source  of  Light  is  Close  to  the  Body.  Let 
the  //  and  V  planes  be  the  receiving  surfaces.  In  Fig.  227,  let 
p  be  any  point  in  space  and  let  /  be  a  source  of  light.  The 
shadow  of  p  is  found  by  passing  a  ray  of  light  through  /  and  ;;. 

This  ray  is  represented  by  the  line  Ip, 
whose  trace  (§31)  falls  on  V,  at  the 
point  s.  The  portion  of  the  ray  of 
light  between  p  and  s  represents  the 
umbra,  and  s  is  the  shadow  of  p,  on  V. 
If  the  source  of  light  is  moved  to  I',  the 
shadow  of  p  falls  on  H,  at  s'. 

II.  The  Sun  Shadow  on  H  or  V. 
If  the  sun  is  taken  as  the  source  of 
light,  the  only  change  from  the  preceding  method  lies  in  the 
manner  of  drawing  the  light  ray.  In  this  case,  each  of  the 
projections  of  the  ray  is  drawn  at  45°  to  HA,  as  shown  by 
the  dot  and  dash  line,  the  shadow  falling  on  H,  at  s". 

207.  Shadow  of  a  Line.  The  shadow  of  a  line  is  determined 
by  the  shadows  of  any  two  of  its  points.  If  the  shadow  falls 
across  HA,  the  problem  lies  partly 
in  a  quadrant  other  than  the  first. 
In  Fig.  228,  let  I  be  a  Source  of 
light,  and  let  it  be  required  to  find 
the  shadow  of  ab.  The  shadow  of 
a  falls  on  H,  at  a\  The  shadow 
of  b  falls  on  V,  at  b\  The  line 
connecting  these  two  points  will 
be  broken  at  HA.     The  direction 

of  the  shadow  line  may  be  established  by  casting  the  shadow 
of  b  as  it  would  fall  on  //  if  it  were  not  intercepted  by  V.  In  the 
figure  this  point  falls  at  b^'.  The  line  a'//'  is  the  shadow  which 
ab  would  cast  if  V  were  a  transparent  plane.  Of  this  line,  the 
portion  a^o  is  the  visible  shadow  on  H,  cast  by  a  part  of  the  given 


XII,  §  209] 


SHADES  AND   SHADOWS 


247 


line  ah.     The  remainder  of  the  Hne  ab  casts  its  shadow  on  V, 

in  the  Hne  b'o. 
208.    Shadow  of  a  Plane  Solid.     The  shadow  of  a  plane  solid 

is  determined  by  the  shadows  of  its  vertices  or  of  its  edges.     In 

Fig.  229,  the  cube  a-h  is  illu- 
minated from  the  point  /.    The 

method  of  casting  the  shadow 

is  found  in  §§  206  and  207. 
It   will    be   noted    that    the 

shadows  of  a  and  g  fall  within 

the  area  put  in  shadow  by  the 

cube.    Thus  the  shadow  of  any 

line  terminating  at  a  or  g,  as 

ad,fg,  etc.,  also  falls  within  the 

shadow  and  does  not  affect  the 

outline   of  the   shadow.     Any 

such    line,    on    the    cube    in 

space,  lies  between  surfaces,  both  of  which  are  in  light  or  in 

shade.     Lines  on  the  cube  which  lie  between  one  surface  which 

is  in  light  and  another  which  is  in  shade,  as  cd,  bf,  etc.,  together 

constitute  the  shade  line  of  the  solid  and  the  shadows  of  these 

lines  together  limit  the  shadow. 

209.  Shadow  FalUng  on  Different 
Planes.  If  the  surface  of  any  solid 
contains  projections  or  recessions 
from  the  main  face,  as  shown  in 
Figs.  230  and  231,  then  the  parts 
farther  back  will  receive,  in  part  or 
in  whole,  the  shadows  cast  by  the 
projections. 

Let  it  be  required  to  cast  the  sun 

shadow  of  the  body  shown  in  Fig.  230.     The  shadow  of  the  line 

bg  will  fall  on  the  surface,  cdef,  while  the  shadow  of  de  falls  on  V. 

The  shadows  of  cb  and  bg  are  determined  in  the  usual  manner 


248 


DESCRIPTIVE   GE0:METRY 


[XII,  §  209 


except  that  the  traces  of  the  light  rays  are  found  on  the  plane 
cdcf,  rather  than  on  H  or  on  V.  It  will  be  noted  that  the  width 
of  the  shadow  1,  is  equal  to  the  depth  of  the  projection  2. 
Also,  and  for  the  same  reason,  3=4  =  5. 

The  preceding  example,  and  the  one  which  follows,  should  be 
studied  carefully  to  fix  in  mind  the  characteristic  shadow  of: 
(1)  a  line  parallel  to  HA  ;  (2)  a  Hne  perpendicular  to  H ;  and 
(3)  a  line  perpendicular  to  F.  Again  notice  that  the  umbra 
formed  by  a  line  is  a  plane,  and  that  the  traces  of  that  plane 

constitute  the  shadow 
of  the  line. 

In  Fig.  231,  a  more 
complex  case  is  shown. 
The  shadow  of  cd  falls 
on  a  surface  perpen- 
dicular to  V,  hence  it 
is  invisible  on  the  ele- 
vation. The  shadow 
of  de  is,  in  general,  a 
continuation  of  a'c% 
but  it  is  a  broken  line, 
owing  to  the  irregular 
surface  on  which  it 
falls.  The  shadow  of 
the  point  h  touches  the  shade  line  mn  at  h%  and  continues,  in 
the  shadow^  of  that  line,  to  h^'.  Similar  cases  occur  at  /',  f 
and  k%  and  in  general,  wherever  a  shade  line  is  crossed  by  a 
shadow.  Such  a  point  is  known  as  a  point  of  loss.  The  line 
f'g  is  a  shade  line,  the  umbra  of  which  coincides  wath  the  sur- 
face gopq ;  hence  this  surface  is  dark.  The  rest  of  the  shadow 
is  easily  found  in  the  usual  way. 
210.   Use  of  the  Profile  Projection. 

It  is  obvious  that  if  the  shadow  of  any  object  is  to  be  deter- 
mined, all  three  dimensions  of  the  object  must  be  known.     In 


Fig.  231 


XII,  §  210] 


SHADES  AND   SHAD(3WS 


249 


general,  two  drawings  are  required  to  give  the  necessary  informa- 
tion. For  the  most  part,  shadows  are  used  in  connection  with 
elevations  or  sections.  The  plan  is  used  merely  to  assist  in  the 
work,  and  the  shadows  on  it  are  not  often  drawn  in.  Sometimes  it 
is  convenient  to  use  three  or  more  drawings,  or  the  shadows  on  all 
drawings  may  be  required.  Again,  the  plan  may  be  discarded 
in  favor  of  a  second  elevation  or  section.  The  particular  method 
selected  for  a  given  problem  is  wholly  a  matter  of  convenience 
and  of  the  desired  result.  The  only  necessary  condition,  com- 
mon to  all  cases,  is  that  the  drawings  used  must  together  fully 


v//////////////////////y//////////////. 


Vy 


Z 


Fig.  232 


describe  all  three  dimensions  of  the  object.  Shadows  cannot 
be  cast  on  an  elevation  without  the  help  of  a  plan,  another 
elevation,  a  section,  or  other  equivalent  data. 

In  Fig.  232  are  shown  two  elevations  of  a  dormer  window 
on  a  steep  roof,  illuminated  from  the  point  L.  The  shadow  will 
fall  on  the  roof,  hence  the  traces  of  the  light  rays  on  the  plane 
of  the  roof  are  required  to  determine  it.  Points  1,  5,  and  6  lie 
in  the  roof  plane,  hence  they  will  coincide  with  their  own 
shadows.  The  shadows  of  the  other  points  are  found  in  the 
order  in  which  they  are  numbered.     Points  5,  7,  and  8  are 


250 


DESCRIPTIVE   GEOMETRY 


[XII,  §  210 


auxiliary  points  used  to  determine  the  directions  of  the  shadows 
of  the  lines  on  which  they  occur,  though  they  are  not  on  the 
outline  of  the  required  shadow. 

211.  The  Shadow  of  a  Circle.  A  circle  illuminated  from  a 
near-by  point  of  light  casts  an  umbra  in  the  form  of  a  frustum 
of  a  cone,  the  apex  of  which  is  the  source  of  light.     The  shadow 


Fig.  233 


of  the  circle  on  any  plane  will  be  the  conic  section  cut  by  that 
plane  from  the  umbra. 

In  Fig.  233,  the  circle  ahd  is  illuminated  from  the  point  p. 
The  plane  of  the  circle  being  parallel  to  V,  the  shadow  on  V  will 
be  a  circle,  since  the  circle  itself  and  its  shadow  are  parallel 
sections  of  the  same  cone.  The  center  of  the  shadow  is  deter- 
mined by  casting  the  shadow  of  o,  which  is  found  at  o\  The 
radius  of  the  shadow  is  determined  by  casting  the  shadow  of 
any  point  on  the  circumference,  as  a.  With  the  radius  o'a" 
the  shadow  on  V  can  be  drawn. 

The  shadow  on  //  is  determined  by  casting  the  shadows  of 


XII,  §  212] 


SHADES  AND   SHADOWS 


251 


several  points  on  the  circumference,  as  bed,  and  drawing  a  curve 
through  the  points  thus  found.  The  precise  nature  of  this 
curve  depends  on  the  relative  positions  of  the  circle  and  the 
source  of  light.  The  shadows  on  V  and  H  will  intersect  in 
HA,  at  X  and  y. 

212.    The  Shadow  of  a  Cone. 

I.  When  the  Source  of  Light  is  Near  the  Cone.  Let 
the  cone  ao  and  the  source  of  light  p  be  as  shown  in  Fig.  234. 
The  rays  of  light  passing  from  p  tangent  to  the  cone,  form 
two  planes  which  pass 
through  p  and  which 
are  tangent  to  the 
cone.  The  intersec- 
tion of  these  planes 
with  H  and  V  will 
give  a  part  of  the 
required  shadow.  The 
balance  of  the  shadow 
is  cast  by  the  base 
circle. 

To  pass  the  planes 
through  p  tangent  to 
the  cone,  an  auxiliary 
line  pa  and  an  auxil- 
iary cutting  plane 
H' A'    were    used,    as 

described  in  §  154.  In  this  manner  the  traces  of  the  planes  X 
and  Y  which  limit  the  shadow  were  found.  The  shadow  of 
the  base  circle  was  found  as  described  in  §  211.  The  shadow 
of  the  base  circle  should  be  tangent  to  the  traces  of  the 
planes. 

The  lines  of  tangency  betWeen  the  cone  and  the  plants  X 
and  Y  (for  example,  ac)  are  the  shade  lines  on  the  cone.  The 
point  c  may  be  established  by  projecting  back  from  c^  to  c^. 


Fig.  234 


252 


DESCRIPTIM^  GE0:METRY 


[XII,  §  213 


II.  The  Sux  Shadow  of  a  Coxe.  In  Fig.  235,  the  sun 
shadow  of  the  cone  base  is  the  circle  6*c*c?V,  and  the  shadow 
of.  the  apex  on  the  same  plane  is  a^  Lines  from  a^\  tangent 
to  the  circle  give  the  H  traces  of  the  shadow  planes.  The 
V  traces  of  the  shadow  planes  pass  through  the  shadow  of  the 

apex.  Let  the  student  determine 
how  the  sun  shadow  of  the  right 
circular  cone,  resting  on  H,  varies 
as  the  angle  of  inclination  of  the 
elements  of  the  cone  varies  between 
0°  and  90°. 

213.  The  Shadow  of  a  Sphere. 
The  shadow  of  a  sphere  cast  by  a 
near-by  source  of  light  can  be  found 
by  passing  a  cone  tangent  to  the 
sphere,  as  described  in  §  173,  and 
then  finding  the  intersection  of  this 
cone  with  the  planes  receiving  the 
shadow.     (See  Fig.  225.) 

If    the    sun    shadow    is    required, 
substitute  for  the  cone  in  the  para- 
graph above  a  cylinder  whose  elements  are  parallel  to  a  light 
ray. 

214.  To  Cast  the  Shadow  of  a  Straight  Line  on  a  Curved 
Surface. 

I.  First  Method.  The  umbra  of  a  straight  line  is  a  plane. 
This  plane  will  cut  the  given  surface  in  a  curved  line,  which  is 
the  required  shadow. 

If  the  source  of  light  is  a  given  point,  pass  a  plane  through 
the  point  and  the  given  line,  as  in  §  69,  and  find  the  inter- 
section of  this  plane  with  the  given  surface,  as  in  §  149.  It 
will  frequently  happen  that  the  traces  of  the  light  plane  do  not 
fall  conveniently  for  this  solution.  In  that  case  the  following 
method  may  be  used. 


Fig.  235 


XII,  §  214] 


SHADES  AND   SHADOWS 


253 


11.  Second  Method.  The  method  given  here  is  analogous 
to  the  shcing  method  of  §  149,  and  furnishes  the  basis  for  the 
solution  of  the  following  two  problems. 

Let  the  hemisphere  and  the  line  ab,  Fig.  236,  be  given,  and 
let  it  be  required  to  find  the  sun  shadow  of  ab  on  the  hemisphere. 
First  cast  the  shadow  of  ab  on  H,  as  a'b'.  Since  the  hemi- 
sphere rests  on  H,  the 
points  c  and  d,  being 
common  to  the  shadow 
and  the  surface,  will 
be  the  terminal  point: 
of  the  required  curve. 

Now  pass  a  cuttin;: 
plane  parallel  to  H,  as 
shown  by  H'A',  and 
cast  the  shadow  of  ah 
on  this  plane,  as  a^'  b'' . 
This  plane  cuts  tht 
circle  men  from  the 
hemisphere.  The  in- 
tersections, e  and  /,  of 
this  circle  with  the 
shadow  a^'b^',  give  two 
more  points  on  the  re- 
quired curve.  Other 
points  are  established  by  means  of  other  cutting  planes. 

The  shade  line  ghk  is  established  by  using  the  projection  on 
a  r'  plane  parallel  to  a  light  ray  as  indicated  at  H" A" .  Using 
this  view,  we  draw  the  light  ray  hs  tangent  to  the  sphere, 
constructing  the  angle  d  as  shown  in  Fig.  226.  This  line  is 
the  top  element  of  a  cylinder  tangent  to  the  sphere  (§  213). 

The  circle  of  tangency  between  the  sphere  and  the  light 
cylinder  is  ghh,  and  its  shadow  in  H  is  gsh.  The  yoint  of  loss 
(§  209),  q,  casts  its  shadow  at  the  intersection  of  a'b'  and  g^s^k^. 


254 


DESCRIPTIVE   GEO:^IETRY 


[XII,  §  215 


215.  To  Cast  the  Shadow  of  a  Curved  Line  on  a  Curved 
Surface.  Let  the  curve  be  made  the  directrix  of  a  cyhnder, 
or  of  a  cone  in  the  case  of  a  near-by  source  of  hght.  This 
cyhnder  or  cone  defines  the  umbra  and  its  intersection  with  the 
given  surface  defines  the  shadow. 

In  Fig.  237,  let  the  hemisphere  o,  and  the  circle  c,  be  given  and 
let  it  be  required  to  find  the  sun  shadow  of  c  on  o.     Light  rays 

tangent  to  c  at  the  usual  in- 
clination define  the  umbra, 
whose  intersection  with  o  is 
determined  as  in  §§  149,  177. 
Vertical  cutting  planes  are 
used  in  this  instance.  In  the 
figure,  one  such  plane  is  drawn, 
the  resulting  points  on  the 
shadow  being  a  and  b. 

The  analysis  used  in  the 
second  method  of  §  214, 
namely  that  of  casting  the 
shadow  of  the  given  line  on 
successive  planes,  may  be  used 
in  this  problem  without  change 
in  the  figure. 

Figure  238  shows  a  niche 
composed  of  a  half  of  a  cylinder 
and  a  quarter  of  a  sphere  and  illuminated  from  the  point  L. 
The  shadow  within  the  niche  is  cast  by  the  line  abed.  The 
shadow  begins  at  the  point  of  tangency,  c,  between  the  curve 
bed  and  a  ray  of  light  through  L.  Succeeding  points  on  the 
shadow  are  found  by  casting  the  shadow  of  bed  on  a  series  of 
vertical  planes,  as  explained  in  §  214.  One  such  plane  is 
drawn.  Here  the  line  cut  from  the  niche  is  fghj  and  the  shadow 
of  abed  on  V  is  khin.  The  point  e  is  the  shadow  of  the  point 
p.     The  line  bg  casts  its  shadow  in  the  straight  line  b'q\ 


Fig.  237 


XII,  §  216] 


SHADES  AND  SHADOWS 


255 


r->in 


216.  The  Shade  and  Shadow  on  a  Double-Curved  Surface  of 
Revolution.  The  shade  Hne,  or  separatrix,  on  a  double-curved 
surface  of  revolution  is  the  line  separating  the  lighted  from  the 
unlighted  portion  of  the  surface,  as  shown  in  Fig.  225,  and  as 
defined  in  §§  203  and  205.  The  cone  of  rays  shown  in  Fig.  225 
is  easily  determined  in  the 
special  case  of  the  sphere,  L' 
as  described  in  §  213.  When 
the  surface  in  question  is 
more  complex,  the  cone  of 
rays  is  not  so  easily  deter- 
mined and  other  methods 
become  necessary. 

Three  methods  are  given 
below,  each  of  which  has 
special  merits,  but  none  of 
which  is  satisfactory  for  every 
case.  All  the  methods  de- 
pend upon  locating  a  large 
number  of  points  of  tangency 

between   straight    lines    and    h 

— « 

curves.  Therefore  the  de- 
gree of  accuracy  for  a  given  surface  depends  upon  choosing  a 
method  which  w^ill  yield  curves  whose  tangents  can  be  readily 
located.  Curves  which  turn  sharply  at  the  point  of  tangency 
are  preferable.  In  most  such  problems,  it  probably  will  be 
necessary  to  use  a  combination  of  two  or  more  of  the  methods 
described  below. 

For  sun  shadows,  special  methods  can  be  devised  which  are 
much  less  laborious  than  any  given  here,  since  the  rays  of 
light  may  be  assumed  to  be  parallel.  These  methods  are 
to  be  found  in  any  standard  text  on  Shades  and  Shadows. 
However,  the  methods  given  here  may  be  adapted  to  the  case 
of  sun  shadows  if  desired. 


Fig.  238 


256 


DESCRIPTIVE   GEOMETRY  XII,  §  216 


i.r 


Fig.  239 


XII,  §  216]  SHADES  AND   SHADOWS  257 

To  facilitate  this  study,  the  student  should  review  §§  212- 
213.  The  problem  of  determining  the  shade  and  shadow  of 
a  cone  is  given  in  §  212,  with  three  important  possible  cases. 
The  shadow  of  a  cylinder  and  that  of  a  sphere  are  treated  in 
§  213.  It  is  also  important  to  realize  that  the  shade  line  on 
any  double-curved  surface  of  revolution  will  be  a  continuous 
curved  line,  whose  shadow  will  determine  the  shadow  of  the 
surface.  As  an  aid  in  visualization,  it  may  be  noted  that  any 
double-curved  surface  of  revolution  may  be  considered  as 
approximated  by  a  series  of  cone  frustums  and  that  the  shade 
lines  on  these  frustums  will  indicate,  in  a  general  way,  how  the 
shade  line  will  appear  on  the  surface  of  revolution. 

I.  The  Envelope  Method.  This  method  is  based  on  the 
slicing  method  of  §  149,  and  on  the  envelope  method  of  §  174. 
It  consists  in  first  casting  the  shadow^,  and  working  back  from 
that  to  the  required  shade  line. 

In  Fig.  239,  let  the  baluster  be  illuminated  from  the  light  /. 
Let  a  series  of  horizontal  cutting  planes,  Q,  R,  S,  etc.,  be  used 
to  determine  a  series  of  circular  sections.  Cast  the  shadows  of 
these  sections  (§  211)  on  the  H  plane,  obtaining  the  circles 
whose  centers  are  q%r%s%t%  etc.  Draw  the  envelope  of  these 
shadows.  This  envelope  outlines  the  shadow  of  the  baluster, 
and  hence  the  shadow  of  its  shade  line. 

In  order  to  determine  the  shade  line  on  the  baluster,  the  points 
of  tangency  between  the  envelope  and  the  shadows  of  the 
circular  sections,  as  a,  b,  c,  etc.,  may  be  used.  Pass  a  light  ray 
through  each  such  point  and  through  /^.  It  will  cut  the  corre- 
sponding circle  on  the  plan  of  the  baluster  in  a  point  which  is 
on  the  line  of  shade,  as  a^',b^,c^.  The  points  m  and  7i,  where  the 
envelope  line  passes  within  the  shadow,  are  points  of  loss  (§  209), 
on  the  baluster. 

This  method  will  serve  quite  well  for  simple  surfaces  of  small 
dimensions.  For  surfaces  of  slight  curvature  it  is  perhaps  the 
best  of  the  three  methods  given  in  this  article. 


258  DESCRIPTIVE   GEOMETRY  [XII,  §  216 

II.  The  Tangent  Cone  Method.  This  method  depends 
upon  the  fact  that  if  a  cone  is  passed  tangent  to  any  double- 
curved  surface,  as  in  Fig.  240,  the  shade  line  on  the  cone,  at  the 
circle  of  tangency,  will  give  a  point  s  on  the  separatrix  of  the 
surface. 

In  Fig.  241,  the  shade  line  on  the  torus  is  determined  in  this 
manner.  The  source  of  light  is  at  /.  Let  the  horizontal  plane 
^yl  be  passed  through  the  torus,  cutting  a  circle  from  the  sur- 
face. Let  this  circle  be  the  base  of  a  cone  which  is  tangent 
to  the  torus  and  whose  apex  is  at  o.  Pass  two  planes,  Q  and  R, 
through  I  and  tangent  to  this  cone  (§  154).      (In  the  figure  RR'' 


/ 

J 

Fig. 

240 

and  QQ  are  the  traces  of  these  planes  on  the  plane  A.)  These 
planes  give  the  shade  lines  oq  and  or,  on  the  cone  (§  212).  The 
points  q  and  r  are  points  on  the  shade  line  of  the  torus. 

For  the  lower  part  of  the  torus,  the  cones  point  downward, 
but  the  solution  is  the  same  in  principle.  Considerable  in- 
genuity is  possible  in  handling  this  method  when  the  apex  of 
a  cone  falls  off  the  paper  or  on  a  level  with  /. 

If  the  shadow  of  the  surface  is  required,  it  may  be  cast  as  in 
I  above,  or  point  by  point  from  the  shade  line,  or  by  a  combina- 
tion of  the  two  methods. 

This  method  gives  a  fair  degree  of  accuracy,  but  it  is  not  well 
adapted  to  cases  like  Fig.  242,  where  the  surface  casts  a  shadow 
on  itself. 


XII,  §  216] 


SHADES  AND  SHADOWS 


259 


Fig.  241 


260 


DESCRIPTIVE  GEOMETRY  [XII,  §  216 


FiQ.  242 


XII,  §  216]  SHADES  AND   SHADOWS  261 

III.  The  Secant  Plane  Method.  In  this  method  use  is 
made  of  secant  planes  passing  through  the  source  of  light  and 
cutting  the  surface  in  a  curved  line,  as  in  Fig.  242.  A  ray  of 
light  in  such  a  plane  and  tangent  to  the  curve  of  intersection, 
r,  s,  t,  will  determine  one  or  more  points,  as  s  or  t,  on  the  separa- 
trix  of  the  surface. 

In  Fig.  242,  the  plane  Z,  passed  through  /,  cuts  the  vase  in 
the  curve  a,  h-f  (§  175).  The  ray  of  light,  Ic,  drawn  tangent 
to  this  curve  determines  the  point  c  on  the  required  separatrix, 
and  also  the  shadow  of  c,  at  c,  on  the  pedestal. 

If  the  shadow  of  the  vase  is  required,  it  may  be  cast  point  by 
point,  or  by  method  I  above. 

This  method  is  a  good  one  when  the  given  surface  is  reentrant, 
and  when  the  curvature  is  not  too  slight. 


CHAPTER   XIII 
OTHER   METHODS    OF   PROJECTION 

217.  Introduction.  Orthographic  projection  makes  use  of 
three  mutually  perpendicular  planes  of  projection  and  of  pro- 
jecting lines  perpendicular  to  these  planes  (§7).  Because  of 
the  three-dimensional  nature  of  the  space  in  which  we  live,  this 
is  the  simplest  and  most  natural  possible  method  of  projection. 
The  relation  of  perpendicularity  between  lines  and  planes  is 
a  unique  relation  and  a  most  useful  one  Again,  most  of  the 
bodies  represented  in  projection  are  largely  composed  of  three 
mutually  perpendicular  sets  of  planes. 

In  §  8  it  was  shown  that  if  it  is  desired  merely  to  represent 
a  body  (as  distinguished  from  locating  it),  one  of  the  planes  of 
projection  may  be  dispensed  with.  Again  in  §  27  and  frequently 
elsewhere,  it  was  shown  that  for  special  purposes  one  of  the 
planes  of  projection  may  be  shifted  at  will.  In  Chapter  XII, 
all  the  work  on  shades  and  shadows  is  done  in  orthographic 
projection,  in  the  usual  manner.  However,  the  shadows  thus 
determined  are,  in  themselves,  projecticns  of  the  given  bodies, 
arrived  at  by  another  method.  In  the  case  of  sun  shadows,  the 
parallel  light  rays  are,  in  effect,  projecting  lines  having  a  special 
relation  to  the  planes  of  projection.  WTien  a  near-by  source 
of  light  is  used,  the  projectors  are  divergent.  Any  surface 
whatever  may  receive  the  projection. 

In  §  221  it  will  be  shown  that  by  placing  an  object  in  a  par- 
ticular position  with  regard  to  the  planes  of  projection,  a  special 
and  very  useful  kind  of  representation  is  obtained. 

In  perspective  projection  the  projectors  diverge  from  a  view- 
point, chosen  at  will,  and  the  projection  is  made  on  a  surface 
between  this  point  and  the  body  which  is  to  be  represented. 

262 


XIII,  §  218]     OTHER   METHODS   OF   PROJECTION       263 


The  surface  on  which  the  projection  is  drawn  is  usually  plane, 
though  sometimes  it  is  cylindrical,  or  (very  rarely)  spherical. 

Various  other  methods  of  projection  have  been  devised  for 
special  purposes,  such  as  for  making  maps  of  various  kinds, 
and  for  military  drawing,  machine  drawing,  and  other  special 
forms  of  drawing.  In  each  case  the  projecting  surfaces  and  lines 
are  so  chosen  as  best  to  explain  the  particular  features  of  the 
object  to  be  represented. 

It  becomes  evident  that  the  position  or  number  of  projecting 
planes,  the  relation  between  the  projecting  planes  and  project- 
ing lines,  or  the  position  of  the  object  to  be  projected,  may  be 
varied  at  will  for  special  purposes.  In  the  present  chapter 
a  number  of  these  special  variations  that  are  most  generally 
used  will  be  discussed. 

218.  One  Plane  Descriptive  Geometry.  Any  point  in  space 
becomes  fixed  when  its  projection  on  one  plane  and  its  distance 


Orthographic.  One   plane. 

Fig.  243 

from  that  plane  are  known.  This  gives  rise  to  a  system  of 
projection  in  which  one  projection,  usually  the  plan,  is  used 
exclusively.  In  this  system  the  elevations  of  the  various  points 
are  indicated  by  index  numbers  on  the  horizontal  projection. 
In  Fig.  243,  the  line  ab  is  shown  in  orthographic  projection 
and  also  in  one  plane  projection.  The  figures  2  and  6,  on  the 
one  plane  projection,  are  the  indexes  of  height. 

The   H  trace  of  ab  may  be  found  by  extending  a'^b^  beyond 
a^  a  distance  equal  to  J  of  its  length. 


264 


DESCRIPTIVE   GEO:\IETRY 


[XIII,  §  218 


Fig.  244 


A  plane  may  be  shown  by  its  H  trace  and  one  other  point  in 
the  phine,  or  by  a  Hne  of  greatest  dechvity  of  the  plane.     Thus 

in  Fig.  244,  the  Hne  oo^  and  the  point 
a^  describe  the  plane  0.  It  can  also 
be  described  by  drawing  a  line  of 
greatest  declivity  (§  90),  a%^. 

Let  it  be  required  to  find  the 
point  where  the  line  d^c^*^,  Fig.  244, 
pierces  the  plane  0.  Extend  d^c^^ 
to  its  trace,  e^,  as  above  described. 
Draw  XX^,  the  trace  of  the  plane  X, 
of  which  c^c^^  is  the  line  of  greatest 
declivity.  Now  the  line  fg^,  drawn  parallel  to  A^Y'',  lies  in  X 
at  the  height  8.  Similarly,  aY  lies  in  0  at  the  height  8.  Hence 
g^  lies  in  both  X  and  0,  as  does  also  h^.  The  line  hY  is  then 
the  line  of  intersection  of  A"  and  0,  and  the  point  j  is  the  trace 
of  d^c^^  on  0.  Other  problems  might  be  worked  out,  but  this 
one  will  sufhce  to  indicate  the  general  nature  of  the  subject. 

The  principles  indicated  above  are  frecjuently  used  in  drawing 
complicated  roof  plans. 
By  this  means  it  is 
often  possible  to  draw 
a  complete  roof  plan 
before  any  accurate 
elevations  have  been 
made.^ 

219.  Lines  of  Level. 
From  the  considera- 
tions of  §  218,  it  follows 
that   a   plane   may  be 

fully  described   by   a  series  of  parallel,  equally  spaced  lines, 
carrying  proper  indexes,  as  shown  for  the  plane  A'  in  Fig.  245. 


1  One  textbook  on  Descriptive  Geometry  uses  the  method  of  one  phine 
projection,  with  modifications,  very  freely. 


XIII,  §  219]     OTHER   METHODS   OF  PROJECTION        265 

These  lines  are  called  the  lines  of  level  of  the  surface.  The 
actual  slope  of  such  a  plane,  with  respect  to  //,  may  be  shown 
by  passing  a  cutting  plane,  Z,  perpendicular  to  the  lines  of 
level.  This  plane  intersects  A^  in  the  line  a%'^.  Now  let  Z  be 
rotated  into  //,  on  ZZ^  as  an  axis.  The  point  a^  is  in  the  axis 
of  rotation,  and  does  not  move,  but  b"^  will  rotate  to  a  point, 
6'^,  distant  7  height  units  from  5^.  The  line  a^6'"  is  now  the 
rabattement  of  a  line  of  greatest  declivity  of  the  plane  X. 
Hence  </>  is  the  true  angle  between  .Y  and  H.     A  section  through 


Fig.  246 


A'  cut  by  any  other  vertical  plane,  as  Y,  can  be  found  in  the 
same  way.  The  line  c^c?'^  shows  such  a  section,  rotated  into  H. 
The  angle  cf)'  is  the  angle  between  the  line  c^(P  and  H. 

Lines  of  level  may  be  used  also  to  describe  planes  intersecting 
in  various  ways,  as  shown  by  the  roof  plan  in  Fig.  194.  It  is 
also  possible,  and  sometimes  convenient,  to  describe  certain 
curved  surfaces  by  means  of  lines  of  level.  Figure  246  shows 
a  surface  composed  of  a  hemisphere  and  mitered  cylinders 
described  by  lines  of  level,  together  with  a  section  cut  from 
the  surface  by  a  vertical  plane  Z.  Without  any  index  of  heights, 
the  surface  shown  may  be  either  convex  or  concave  toward  the 
observer.     A  set  of  height  indexes  w^ould  make  this  definite. 


266 


DESCRIPTIVE   GEOMETRY 


[XIII,  §  220 


220.  Topography.  ^Yhen  it  becomes  necessary  to  describe 
accurately  a  very  irregular  surface,  such  as  a  piece  of  land,  lines 
of  level  are  found  very  useful.  A  map  on  which  lines  of  level 
are  shown  is  called  a  topographical  map,  or  a  contour  map,  and 
the  lines  of  level  are  usually  spoken  of  as  contour  lines.  Such 
maps  may  be  drawn,  of  course,  at  any  scale,  and  the  intervals 
of  height  between  contour  lines  may  be  chosen  to  suit  the 
particular   case,   the  vertical    scale    commonly    being  different 


Fig.  247 

from  the  horizontal  scale.  Shorter  intervals  between  the 
contour  lines  give  greater  accuracy,  but  they  involve  more 
labor,  of  course. 

The  natural  surface  of  the  earth  follows  no  geometrical  law. 
Hence  its  lines  of  level  are  free-hand  curves  (§  108).  But 
where  land  has  been  graded  into  formal  terraces,  roads,  etc., 
the  surfaces  become  geometric  and  the  contour  lines  may  be 
straight  lines  or  regular  curves.  Original  plans  for  grading  are 
indicated  in  this  manner. 


XIII,  §  2201     OTHER   METHODS  OF   PROJECTION       267 


Figure  247  shows  a  contour  map  which  includes  a  lake,  a 
stream,  and  two  hills.  Contours  are  indicated  at  five-foot 
intervals,  starting  from  the  lake  level  as  the  zero  datum.  The 
contour  of  the  lake  bottom  is  shown  by  dotted  lines.  Hill  60  is 
seen  to  be  much  steeper  on  the  northwest  face  than  elsewhere  and 
drops  toward  the  lake  in  a  concave  spur,  while  hill  47  descends 
on  the  east  into  the  ravine  through  whirh  the  stream  flows. 


Fig.  248 

In  Fig.  248  the  same  piece  of  land  is  shown  as  in  Fig.  247, 
but  it  has  now  been  graded  so  as  to  give  a  flat  space  at  level  25. 
The  front  part  of  this  space  is  terraced  up  while  at  the  back 
it  is  cut  into  the  hills. 

In  making  contour  maps,  the  engineer  starts  from  some 
fixed  and  known  point  as  a  bench-mark,  and  determines  by 
stadia  ^  the  bearing,  distance,  and  levels  with  reference  to  the 

^  By  means  of  the  stadia  it  is  possible  to  determine  the  distance  and 
direction  between  two  points,  as  well  as  their  relative  levels,  at  one  setting 
of  the  instrument. 


268 


DESCRIPTIVE   GEOMETRY  [XIII,  §  220 


known  point  of  as  many  points  as  can  readily  be  seen  from  this 
position.  The  instrument  is  then  moved  to  a  new  known 
position,  and  other  points,  hidden  before  but  now  visible,  are 
located.  From  the  data  thus  secured,  the  points  are  mapped 
and  their  hei^fht  indexes  are  recorded.     Contour  lines  are  then 


Fig.  249 


drawn  through  corresponding  heights,  interpolation  being  used 
freely. 

As  an  example  of  the  type  o"  problem  that  arises  in  contour 
work,  let  the  contours  in  Fig.  249  be  given.  Let  it  be  required 
to  show  the  cut  and  fill  necessary  to  carry  a  level  road  along 
the  face  of  the  hill  as  shown,  assuming  the  angle  of  repose  for 
the  embankments  to  be  30°.  The  road  follows  the  25  ft.  con- 
tour closely.  Henor,  if  the  surface  of  the  road  is  to  be  level,  it 
will  cut  into  the  hill  on  one  side  and  will  need  embankment  on 


XIII,  §  221]     OTHER  METHODS  OF  PROJECTION       269 

the  other.  The  faces  of  the  cut  and  fill  will  be  planes  where 
the  road  is  straight,  and  conical  where  the  road  curves.  The 
angle  between  the  planes  or  the  cones  and  the  surface  of  the 
road  will  be  30°.  It  is  required  to  find  the  line  in  which  these 
surfaces  will  meet  the  natural  surface  of  the  ground  as  shown 
by  the  original  contours. 

Let  the  surface  of  the  road  at  elevation  25  be  taken  as  H. 
Pass  the  vertical  plane  Z  perpendicular  to  the  axis  of  the  road. 
This  plane  will  be  cut  by  the  planes  of  the  cut  and  fill,  in  lines 
which  make  angles  of  30°  with  H.  Let  these  lines  be  rotated 
about  ZZ^,  into  H.  They  will  appear  at  ahcd.  Locate  e/fg'h'f, 
on  a-d,  at  10'  height  intervals,  above  and  below  the  road. 
These  points,  rotated  back  into  Z,  give  the  points  e-j,  which  lie 
in  the  planes  of  cut  and  fill  and  at  the  same  level  as  the  given 

contours. 

Through  these  points  draw  Hues  of  level,  hh'h'\  gg'g"t  etc., 
in  the  surfaces  of  cut  and  fill.  The  points  in  which  these  lines 
intersect  contour  lines  of  a  corresponding  height,  as  1-12, 
define  the  limits  of  the  cut  and  fill. 

It  should  be  noticed  that  above  the  road  the  natural  contours 
are  cut  away,  while  below  the  road  they  are  covered  under. 

Problems  involving  topography  are  simple  in  principle. 
The  beginner's  errors  usually  include  streams  which  flow  uphill 
or  along  the  crest  of  a  ridge,  lakes  which  are  more  or  less  out 
of  level,  or  terraces  which  intersect  in  impossible  lines.  These 
errors  can  be  avoided  only  by  close  attention  and  clear  visual- 
ization. 

221.  Isometric  Projection.  In  isometric  projection  (§§  2, 
217)  the  projecting  fines  are  drawn  perpendicular  to  the  plane 
of  projection,  but  the  body  to  be  projected  is  placed  in  a 
special  position,  in  order  to  produce  a  drawing  which  has 
some  of  the  pictorial  qualities  of  a  perspective,  but  which 
can  be  more  easily  made,  and  which  has  some  value  as  a 
working  drawing. 


270 


DESCRIPTIVE  GEOMETRY         [XIII,  §  221 


If  a  cube  is  so  placed  in  orthographic  projection  that  each  of 
the  edges  which  meet  at  a  given  corner  make  equal  angles  with 
the  V  plane  (Fig.  250)  it  is  obvious  that  three  of  its  six  faces 
become  visible  in  projection.  Moreover,  the  projections  of  the 
angles,  aoc,  aoe  and  coc,  are  equal,  and  each  is  equal  to  120°. 
Again,  all  of  the  edges  are  equal  in  projection,  but  the  length  of 
the  projections  is  less  than  the  true  length  of  the  edges  of  the 
cube. 

On  the  other  hand,  the  face  diagonals,  ac,  ae,  and  ec,  are 
in  a  plane  parallel  to  V,  and  hence  they  are  projected  in  their 
true  length.  Finally,  the  diagonals  ob,  od,  of,  are  lines  of 
greatest  declivity  of  their  respective  planes, 
and  hence  they  are  foreshortened  more  than 
the  edges. ^ 

It  is  now  evident  that  any  object  com- 
posed of  three  sets  of  mutually  perpendicu- 
lar planes  may  be  projected  by  drawing  one 
st't  of  edges  vertically,  while  the  others 
make  angles  of  30°  with  the  horizontal,  and 
laying  these  edges  off  at  a  reduced  scale. 
Now  let  the  scale  at  which  the  drawing  is  made  be  arbitrarily 
increased  (about  22%)  so  that  all  main  edge  lines  and  all  lines 
parallel  to  them  are  drawn  in  their  true  length.  The  resulting 
drawing  is  an  isometric  projection.  On  it  all  lines  parallel 
to  the  main  edges  (called  isometric  lines)  are  shown  in  their  true 
length,  and  either  vertically  or  at  an  angle  of  30°  to  //.  All 
other  lines  are  distorted  in  length ;  some  are  reduced  in  size, 
while  others  are  magnified.  Fortunately,  the  lines  that  are  of 
natural  length  occur  most  frequenth'. 

1  Let  the  true  length  of  any  side  of  the  above  cube  be  S.  Then  the  true 
length  of  a  diagonal  is  S\/2.  Now  remembering  that  the  angle  aoc  =  120°, 
oab  =  60°,  etc.,  prove  that  the  projected  length  of  any  side  of  the  cube  is 
equal  to  0.81G47.S.  The  ratio,  .816+,  is  called  the  isometric  scale,  and  it 
represents  the  reduction  in  size  of  the  edges  of  the  cube.  It  may  be  shown 
also  that  for  lines  not  parallel  to  the  edges,  the  scale  will  vary  from  0.5773 
(for  bo,  of,  od)  to  1.000  (for  ac,  ce.  ej). 


XIII,  §  222]    OTHER  ^METHODS   OF  PROJECTION       271 

222.  Typical  Case.  Figure  251  shows  the  elevation  and 
section  of  a  flat  arch  in  a  battered  wall.  Let  it  be  required  to 
show  the  keystone  in  isometric  projection.  For  the  sake  of 
clearness,  the  scale  of  the  detail  is  enlarged.  First  draw  around 
the  given  stone  a  parallelepiped  which  incloses  it  and  coincides 
with  its  extreme  faces,^  as  shown  by  a-h.  Draw  this  block 
in  isometric  projection,  as  a'-h',  assuming  a  point  of  view 
above,  in  front,  and  to  the  right.     The  points  j'k'l'm'  are  found 


Fig.  251 

by  measuring  along  the  isometric  lines  of  the  block  distances 
determined  from  the  section  and  elevation.  Any  point  w^ithin 
the  block,  as  z,  may  be  located  by  measuring  along  isometric 
lines,  as  h'x,  xy,  yz' ,  from  some  known  starting  point.  The  dis- 
tance between  any  two  points  can  be  found  in  a  similar  manner. 
Figure  251  (2)  shows  an  isometric  projection  of  the  same  stone 
in  a  different  position.  Here  the  bottom,  rear,  and  left-hand 
faces  appear  as  visible. 

1  The  French  call  this  inclosing  block  the  solide  capable;    a  term  which 
if  translated  literally  gives  a  very  good  notion  of  its  nature  and  use. 


272 


DESCRIPTIVE   GE0:METRY 


fXIII,  §  223 


223.  Curves  in  Isometric  Projection.  The  general  method 
for  drawing  curves  in  isometric  projection  consists  in  making 
an  isometric  drawing  of  a  polygon  w^hich  may  be  inscribed  in  or 
circumscribed  about  the  given  curve,  and  using  this  polygon 
as  a  guide  in  tracing  the  curve,  as  illustrated  in  Fig.  252  a. 

A  circle  in  isometric  projection  appears  as  an  ellipse.  It 
may  be  most  easily  drawn  by  use  of  a  circumscribed  square,  as 
in  Figure  252  b.  The  diameters,  ab  and  cd,  of  the  square  are 
conjugate  diameters  (§§121,  124,  II)  of  the  ellipse,  which  may  be 
constructed    as    shown    in    Fig.    134.     Another    construction, 


Fig.  252  a 

which  approximates  the  required  ellipse  by  arcs  of  a  circle, 
(§  110,  III)  may  be  made  as  follows.  In  Fig.  252  b  draw  ea,  ec, 
fdy  and  fo.  From  c  as  a  center  swing  the  arc  ac  and  from  h  as 
a  center  the  arc  cb.  The  other  half  of  the  curve  has  its  centers 
at/  and  g. 

224.  Curved  surfaces  in  Isometric  Projection.  The  iso- 
metric projection  of  a  cone  can  be  established  by  drawing  the 
base,  the  apex,  and  two  lines  from  the  apex  tangent  to  the  base. 

The  isometric  projection  of  a  cylinder  is  given  by  two  bases 
with  their  common  tangents. 

Since  a  sphere,  in  any  position  whatever,  projects  as  a  circle, 
its  isometric  projection  will  be  a  circle,  exaggerated  in  size  in 
the  ratio  122%.  The .  isometric  projection  of  any  double- 
curved  surface  may  be  made  by  the  Envelope  Method  (§  174). 


XIII,  §  225]     OTHER  METHODS   OF  PROJECTION        273 


Warped  surfaces  are  best  shown  in  isometric  projection  by 
drawing  isometric  projections  of  the  directrices  and  of  a  number 
of  elements. 

225.  Oblique  or  Clinographic  Projection.  In  oblique  pro- 
jection, the  object  is  placed  with  its  main  faces  parallel  to  a 
plane  of  projection  and  is  projected  on  that  plane  by  means  of 
lines  inclined  in  any  convenient  manner  so  as  to  give  the  desired 
result.  It  will  be  recognized  that  a  sun  shadow  is  a  special  kind 
of  oblique  projection  of  the  object  casting  the  shadow. 

The  plane  of  projection  is  usu- 
ally, though  not  necessarily,  verti- 
cal. In  the  following  description 
it  will  be  so  assumed  and  will  be 
referred  to  as  V. 

It  can  be  seen  readily  that  the 
oblique  projection  of  any  face  of 
an  object  that  is  parallel  to  V  will 
be  the  true  size  and  shape  of  that 
face.  Also  the  projection  of  any 
line  perpendicular  to  V  will  vary 
as  to  length  with  the  inclination  of 
the  projecting  lines  toward  V,  and 
will  vary  as  to  slope  with  the  in- 
clination   of    the    projecting    lines 

toward  H.  Moreover,  in  projecting  a  line  which  is  perpen- 
dicular to  V,  the  slope  of  the  projection  and  its  length  may 
be  made  to  have  an^^  desired  values  whatever  by  properly 
selecting  the  inclination  of  the  projecting  lines. 

In  practice  it  is  usually  convenient  to  draw  an  oblique  pro- 
jection without  considering  particularly  the  exact  position  of 
the  projecting  lines.  A  front  elevation  of  the  object  is  drawn 
as  shown  by  the  dotted  lines  in  Fig.  253.  Next  any  con- 
venient slope  is  chosen  for  the  lines  perpendicular  to  the  front 
face.     Thirty  degrees  is  ordinarily  satisfactory,  and  is  the  slope 


Fig.  253 


274  DESCRIPTIVE  GEOMETRY  [XIII,  §  225 

used  in  the  figure.  In  laying  off  distances  on  these  lines,  any 
proportion  to  the  true  dimensions  may  be  used,  depending  on 
how  fully  it  is  desired  to  show  the  side  faces.  In  the  figure, 
a  one  to  one  ratio  was  used. 

The  inclination  and  length  ratio  to  be  used  for  lines  perpen- 
dicular to  V,  having  been  chosen,  and  the  main  face  of  the  solide 
capable  (§  222)  having  been  drawn,  the  projection  is  con- 
structed by  measurement  along  lines  parallel  to  the  edges  of  the 
solid,  as  explained  in  §  222. 

In  such  a  projection,  the  side  faces  can  be  exaggerated  or 
suppressed  by  varying  the  slope  and  length  ratio  as  noted  above. 
But  it  should  be  noticed  that  the  length  of  diagonal  lines 
is  distorted  (as  in  isometric  projection,  §  221),  and  that  if 
the  exaggeration  of  the  side  faces  is  carried  too  far,  it  is  apt  to 
produce  an  unsatisfactory  drawing.  Figure  253  is  drawn  with 
a  slope  of  30°  and  a  length  ratio  of  unity.  This  particular  case 
is  called  a  Cavalier  Projection  and  is  produced  by  projecting 
lines  making  an  angle  of  45°  with  V.  It  will  not  ordinarily 
be  wise  to  use  either  a  greater  slope  or  ratio  than  the  one  here 
indicated. 


APPENDIX 

The  exercises  on  the  following  pages  may  be  divided  into  two 
parts.  Numbers  I  a  to  LXXIV  b  are  parallel  with  the  similarly 
numbered  exercises  in  the  body  of  the  text  of  Chapters  I  to 
VIII.  The  rest  of  the  problems  are  designed  to  accompany  the 
text  of  Chapters  IX  to  XIV.  These  exercises  employ  the  same 
principles  as  are  explained  in  the  text,  with  sufficient  variation 
of,  the  conditions  to  make  the  solutions  worth  while. 

Sheets  XXII  a  to  XXXVII  a  consist  of  three  problems  each, 
so  spaced  that  number  1  from  any  sheet  may  be  combined  with 
number  2  and  number  3  from  any  other  sheet  or  sheets.  The 
typical  layout  for  exercise  sheets,  applying  to  all  cases  (except 
where  otherwise  specially  noted),  is  given  on  p.  xii  of  the  general 
explanations.  For  Exercise  sheets  I  a  to  XXI  a  the  location  of 
the  profile  plane  is  given  with  the  statement  of  each  set  of 
exercises.  For  all  other  sheets,  no  profile  projection  is  required, 
and  P  is  taken  at  the  right  border  line,  unless  otherwise  specially 
mentioned. 

EXERCISE   SHEET  la    (§13) 

[Note.  For  the  typical  layout  for  exercise  sheets,  applying  to  all 
exercises  (excepting  a  few  which  are  separately  described),  see  p.  xii 
of  the  general  explanations.] 

Take  P  at  the  right,  6  from  the  border  line,  turned  toward  the  given 
points,  into  H.     Draw  all  three  projections  of  the  following  points. 


a 

(68,  11,  16) 

/ 

(35, 

7,13) 

b 

(61,    0,  11) 

9 

(27, 

0,    5) 

c 

(55,    3,  11) 

h 

(19, 

8,    8) 

d 

(48,  17,    5) 

i 

(12, 

12,    9) 

e 

(43,    0,    0) 

3 

(  4, 

7,    0) 

275 


276  DESCRIPTIVE  GEOMETRY 

EXERCISE   SHEET  II  a    (§13) 

Take  P  at  the  left,.  20  from  the  border  Une,  turned  away  from  the 

given  points,   into   V.     Draw  all  three  projections  of  the  following 

points. 

a     (  4,  10,    8)  /     (33,  10,  13) 

b  (  9,    4,  16)  g  (39,    0,    6) 

c  (17,    0,    0)  h  (44,  18,  20) 

d  (23,  13,  13)  i  (50,    8,    2) 

e  (29,  18,    0)  j  (56,  13,  13) 

EXERCISE   SHEET  Ilia    (§13) 

Take  P  at  the  left,  20  from  the  border  line,  turned  away  from  the 

given  points,   into  H.     Draw  all  three  projections  of  the  following 

points. 

a     (  3,    5,    4)  /     (31,    8,  15) 

b  (  9,  12,    2)  9  (37,  15,  0) 

c  (15,  10,    9)  h  (45,  10,  15) 

d  (22,  19,  20)  ^  (50,  16,  12) 

e  (27,    0,    0)  j  (56,    0,  18) 

EXERCISE   SHEET  IV  a   (§13) 

Take  P  at  the  left,  2  from  the  border  line,  turned  toward  the  given 
points,  into  V.     Draw  all  three  projections  of  the  following  points. 

a  (23,  8,  20).  Move  the  point  9  toward  V  and  5  toward  H.  Call 
this  position  a'. 

b  is  29  from  P,  4  farther  from  H,  and  15  nearer  V  than  is  a.  ]Move 
b  downward  5,  forward  10,  and  6  to  the  right.  Call  these  successive 
locations  b',  b",  b'". 

c  is  43  from  P,  11  from  V,  and  on  a  plane  passing  through  HA  and 
making  an  angle  of  60°  with  H. 

d  lies  in  the  same  line  perpendicular  to  P  as  c,  but  10  farther  from 
P.  With  c  as  a  center,  rotate  d  until  it  lies  directly  under  c,  keeping 
d  always  the  same  distance  from  V.     Mark  the  new  position  d'. 

e  lies  on  V,  15  from  //  and  60  from  P. 

/  lies  in  the  same  horizontal  plane  as  c,  1|"  (actual)  from  e  and  7 
in  front  of  V. 

EXERCISE   SHEET  Va   (§20) 

Take  P  at  the  right,  2  from  the  border,  and  opened  toward  the 
given  lines,  into  V.     Draw  the  projections  of  the  following  lines. 


APPENDIX 


277 


Line  ab :  a  (75,  16,  12) ;   6  (06,  6,  3). 

Line  cd  :  cd  is  perpendicular  to  // ;  d  is  at  (62,  0,  9). 

Line  ef :  e  (57,  0,  0) ;  ef  runs  upward  and  to  the  right,  /  being  48 
from  P. 

Line  gh  :  parallel  to  HA,  17  from  V  and  9  from  H ;  ^  is  44  from  P, 
and  h  is  34  from  P. 

Line  ij :  is  in  H,  between  39  and  0  from  P ;  running  forward  and 
to  the  right  from  i. 

Line  kl :  parallel  to  P;  runs  'ownward  and  forward  from  k  (27, 
17,  2). 

Line  mn :  intersects  F  at  w  (23,  20,  0),  and  makes  an  angle  of  45° 
with  F.     n  and  m  are  in  the  same  horizontal  plane. 


EXERCISE   Via    (§21) 
L  escribe  in  writing  the  following  lines  : 

r 


EXERCISE   SHEET   Vila    (§22) 

Take  P  at  the  right,  18  from  the  border  line,  and  turned  away  from 
the  objects,  into  F.  Profile  projections  are  required  for  #3  only.  Use 
notation  carefully. 

1.  The  center  of  a  square,  abed,  10  units  on  a  side,  is  at  e  (50,  10, 
10).  The  plane  of  the  figure  is  parallel  to  H  and  its  sides  make  angles 
of  45°  with  F.  The  corner  a  is  farthest  from  P,  c  is  nearest  P,  and  b 
is  nearest  F.  Rotate  the  square  through  45°,  on  bd  as  an  axis,  a  going 
down  and  c  up.     Draw  its  projections  in  this  position. 

2.  A  circle,  14  in  diameter,  has  its  center  at  o  (31,  10,  10).  It  lies 
in  a  plane  parallel  to  F.  Call  the  horizontal  diameter  fg.  Rotate 
the  figure  on  its  vertical  axis,  through  45°,  and  draw  its  projections. 

[Note.     Assume  several  points  on  the  circumference  as  auxiliaries.] 

3.  Use  the  profile  plane.  A  regular  hexagon  jklnmp  has  its  center 
at  s  (12,  10,  10).     The  diameter  jm  is  parallel  to  HA  and  the  plane 


278  DESCRIPTIVE  GEO:\IETRY 

of  the  figure  is  parallel  to  V.  Draw  its  projections  in  this  position, 
then  rotate  it  on  jm  as  an  axis  lentil  it  lies  in  a  plane  making  45°  with 
H.     Draw  its  projections. 

EXERCISE   SHEET  Villa   (§26) 

Take  P  at  the  left,  16  from  the  border  line,  and  turned  away  from 
the  objects,  into  V.  Draw  the  profile  projections  in  Exs.  2  and  3 
only.     Use  the  first  method. 

1.  Given  a  (3,  16,  4)  and  6  (10,  9,  9).  Find  the  true  length  of  ah 
and  the  true  angle  it  makes  with  H.     Swing  about  b. 

2.  Given  c  (12,  4,  5)  and  d  (19,  13,  10).  Find  the  true  length  of 
cd  and  the  true  angle  it  makes  with  V,  by  swinging  about  c  until  the 
line  is  parallel  to  P. 

3.  Given  e  (23,  11,  6)  and/  (23,  20,  11).  Draw  all  three  projections 
of  ef  and  find  the  true  length  of  ef  and  the  angles  it  makes  with  V  and 
P,  by  swinging  the  line  about  /  as  a  center. 

4.  Given  g  (36,  4,  8)  and  h  (45,  10,  13).  Keeping  h  stationary,  find 
the  true  length  of  gh  and  the  angle  it  makes  with  H.  Check  the  true 
length  by  swinging  the  line  parallel  to  H.     Indicate  the  angle  with  V. 

5.  Given  i  (50,  5,  7),  j  (56,  11,  14),  and  k  (60,  4,  5).  Draw  the 
projections  of  the  triangle  ijk  and  find  its  true  shape. 

[Note.  Find  the  true  lengths  of  the  sides  individually  and  construct 
the  triangle.] 

EXERCISE   SHEET  IX  a   (§27) 

Take  P  at  the  left,  16  from  the  border  line,  and  turned  away  from 
the  objects,  into  H.  Use  P  in  problem  4  only.  Solve  by  the  second 
method. 

1.  Given  a  (13,  12,  8)  and  b  (21,  6,  11).  Find  the  true  length  of 
ab  and  the  angle  it  makes  with  H. 

2.  Given  c  (25,  5,  15)  and  d  (33,  8,  7).  Find  the  true  length  of 
cd  and  the  angle  it  makes  with  V. 

3.  Given  e  (36,  15,  8),  /  (39,  12,  13),  g  (43,  8,  2),  and  h  (45,  6,  6). 
Find  the  true  shape  of  the  quadrilateral. 

4.  Given  i  (4,  4,  3)  and  i  (10,  14,  12).  Find  the  true  length  of  ij 
and  the  angle  it  makes  with  P. 

5.  Given  k  (50,  5,  8)  and  I  (59,  9,  13).  Find  the  true  length  of  kl 
and  the  angle  it  makes  with  H.  Use  an  auxiliary  plane  through  the 
H  projection  of  kl. 


APPENDIX  279 

EXERCISE   SHEET   Xa   (§29) 

Take  P  at  the  right  border  line.     No  P  projections  are  required. 

1.  The  point  a  (76,  5,  4)  is  given.  The  hne  ab  is  10  units  long  and 
slopes  upward,  forward,  and  toward  P  from  a,  making  angles  of  30° 
and  45°  with  H  and  V  respectively.     Draw  its  projections. 

2.  The  point  c  (55,  4,  7)  is  given.  The  line  cd  is  8  units  long,  and 
slopes  upward  and  forward  from  c,  making  an  angle  of  45°  with  H  and 
an  angle  of  30"  with  V.     Draw  its  projections.     (^Two  solutions.) 

3.  The  point  e  (40,  15,  17)  is  given.  The  hne  ef  is  8  units  long  and 
makes  angles  of  45°  and  15°  with  H  and  V,  respectively.  Draw  its 
projections.     (Four  solutions.) 

4.  The  point  g  (27,  4,  5)  is  given.  The  hne  gh  is  12  units  long  and 
makes  angles  of  45°  with  H  and  V,  sloping  upward  and  forvv^ard.  Draw 
its  projections. 

5.  The  point  i  (14,  4,  5)  is  given.  Draw  the  projections  of  the  hne 
ij  12  units  long  and  making  an  angle  of  45°  with  H  and  60°  with  V. 
Explain  the  result. 

EXERCISE   SHEET   XI  a    (§30) 

Take  P  at  the  right  border  line.     No  P  projections  are  required. 

1.  Given  a  (75,  12,  13)  and  b  (66,  6,  9),  locate  c  on  ab  7  units  from  a. 

2.  Given  d  (60,  5,  7)  and  e  (51,  14,  13),  locate  /  on  de  1"  from  e. 
Solve  in  two  different  ways,  and  check  the  results. 

3.  From  g  (32,  8,  5)  the  hne  gh  slopes  away  from  H,  V,  and  P, 
making  an  angle  of  45°  with  H  and  30°  with  V.  The  point  i  lies  on 
9h  36  from  P.     Find  the  length  of  hi. 

4.  Bisect  the  hne  joining  the  pomts  j  (26,  13,  8)  and  k  (18,  9,  13), 
and  mark  the  middle  point  I.  Notice  the  relation  between  segments 
of  projections. 

5.  Given  m  (13,  7,  10)  and  n  (4,  13,  19),  trisect  77in  at  o  and  p. 
State  the  ratios  between  segments  of  the  projections  and  those  of  the 
true  length. 

EXERCISE   SHEET   XII  a   (§  31) 

Take  P  at  the  right,  16  from  the  border  line,  and  turned  away  from 
the  objects,  into  V.     Use  P  only  when  necessary. 

1.  Given  a  (59,  12,  2)  and  b  (51,  3,  9),  find  the  H  and  T'  traces  of  ah, 

2.  Given  c  (46,  11,  6)  and  d  (38,  2,  6),  find  the  traces  of  cd. 

3.  Given  e  (33,  17,  9)  and/  (24,  17,  9),  find  the  traces  of  ef. 

4.  Given  g  (20,  12,  12)  and  h  (20,  4,  4),  find  the  traces  of  gh. 

5.  Given  i  (12,  5,  6)  and  j  (4,  16,  10),  find  the  traces  of  ij. 


280  DESCRIPTIVE  GEOMETRY 

EXERCISE   SHEET   XIII  a   (§33) 

Take  P  at  the  left,  16  from  the  border  Une,  and  turned  away  from 
the  objects,  into  V.     Use  P  only  when  necessary. 

1.  From  a,  the  line  ab  slopes  away  from  P  and  V  but  towards  H. 
From  c  the  line  cd  slopes  away  from  P  and  H  but  towards  V.  These  lines 
intersect  at  e,  which  is  7  from  H  and  6  from  V.  Draw  the  projections 
between  3  and  13  from  P.    (Use  the  P  projections  as  a  test  of  accuracy.) 

2.  Given/  (16,  2,  7),  g  (26,  8,  12),  h  (17,  9,  9),  and  i  (25,  3,  5).  Do 
fg  and  hi  intersect? 

3.  Given  j  (30,  5,  7)  and  h  (39,  10,  13),  draw  7nn  intersecting  jk 
at  I,  which  is  7  units  from  k.  From  m,  mn  slopes  away  from  P  and 
toward  V  and  H. 

4.  Given  p  (47,  5,  7),  q  (58,  11,  14),  and  r  (50,  12,  13),  draw  ro 
intersecting  pq  at  its  middle  point  o. 

EXERCISE    SHEET    XIV  a    (§34) 

Take  P  at  the  left  border  line  and  turned  toward  the  objects,  into  V. 

1.  Draw  all  three  projections  of  the  line  joining  a  (18,  14,  9)  and 
h  (27,  7,  5)  and  those  of  another  line  cd  which  is  parallel  to  ah  but  nearer 
H  and  farther  from  V  than  is  ah. 

2.  Given  e  (31,  10,  4)  and/  (41,  9,  10),  draw  the  projections  of  gh 
which  passes  through  i  (34,  13,  12)  and  is  parallel  to  e/.    Find  its  V  trace. 

3.  Through  I,  the  middle  point  of  the  line  joining  /  (46,  6,  8)  and 
k  (58,  14,  13),  draw  nm  parallel  to  ef,  and  1"  long. 

4.  Through  o,  70  from  P,  draw  op  and  oq  parallel  to  ah  and  jk  re- 
spectively.    Find  the  true  size  of  the  angle  poq.     (See  Note,  p.  40.) 

EXERCISE   SHEET   XV  a    (§§38-42) 

Take  P  at  the  left,  16  from  the  border  line,  and  turned  toward  the 
objects,  into  V} 

1.  Draw  all  three  projections  of  the  following  points. 

a   (16,  16,  9)     h  (II,  20,  10,  7)     c   (III,  24,  10,  13)     d   (IV,  28,  20,  3) 
e   (III,  32,  20,  10)  /  (36,  4,  12)  g   (II,  40,  22,  2) 

2.  Given  the  point  h  (IV,  45,  6,  15),  move  h  parallel  to  HA  until 
it  is  5  farther  from  P.     From  here  move  it  straight  up  15. 

3.  Given  the  point  i  (IV,  60,  12,  3),  move  i  straight  towards  point 
a  of  Ex.  1,  till  it  is  5  nearer  to  P. 

*  In  describing  the  movement  of  P,  the  description  will  be  understood 
to  apply  to  the  movement  of  that  part  of  P  which  lies  in  the  first  quadrant, 
as  heretofore. 


APPENDIX  281 

EXERCISE   SHEET   XVI  a    (§43) 

Take  P  at  the  right,  16  from  the  border  hne  and  turned  away  from 
the  objects,  into  H.     Draw  all  three  projections  of  the  following  hnes. 

1.  Given  a  (II,  61,  4,  6)  and  b  (IV,  53,  8,  3),  find  the  traces  of  ab, 
and  designate  the  quadrants  through  which  it  passes, 

2.  In  the  space  40-50  from  P  draw  the  projections  of  a  line  cd 
running  from  a  point  c  in  I,  through  II,  to  a  point  d  in  III,  and  desig- 
nate the  parts  in  the  different  quadrants.     Do  not  use  the  profile. 

3.  From  e  (IV,  36,  8,  12)  draw  ef,  through  I  into  II,  making  ef 
parallel  to  P.     Indicate  the  parts  that  he  in  I,  II,  and  IV. 

4.  Given  g  (II,  30,  5,  14)  and  h  (II.  18,  9,  3),  from  j  (28,  4,  10) 
draw  ji  to  i,  the  middle  point  of  gh.  Find  the  true  distance  between  j 
and  the  V  trace  of  ji. 

EXERCISE   SHEET   XVII  a   (§44) 

Take  P  at  the  left,  16  from  the  border  line,  and  turned  toward  the 
objects,  into  V. 

1.  Given  a  (II,  4,  9,  7)  and  b  (IV,  12,  4,  10),  draw  all  three  projec- 
tions, and  find  the  H,  V,  and  P  traces  of  ab. 

2.  Through  d  (III,  16,  13,  8)  draw  cd  12  units  long,  making  an  angle 
of  45°  with  H  and  an  angle  of  30°  with  V.  From  d,  cd  slopes  away 
from  P  and  V,  toward  H. 

3.  Given  e  (IV,  31,  4,  15),  /  (IV,  40,  7,  10).  Find  the  true  length 
of  ef  and  the  true  angle  which  it  makes  with  H,  by  the  second  method. 

4.  Through  g  (II,  58,  3,  13)  draw  gh  parallel  to  cd  and  gi  parallel 
to  ab.     Both  h  and  i  are  in  III.     Find  the  true  angle  between  gh  and  gi. 

EXERCISE   SHEET   XVIII  a    (§§46-^8) 

Take  P  at  the  left  border  line.     No  P  projections  are  required. 

1.  Locate  the  point  a  (18,  0,  3).  Draw  ab  in  H,  1"  long,  forward 
and  to  the  right  from  a  and  making  an  angle  of  45°  with  HA.  This 
line  is  one  side  of  a  regular  hexagon  whose  plane  makes  an  angle  of 
45°  with  H  and  slopes  away  from  V.     Draw  its  H  and  V  projections. 

2.  Draw  the  projections  of  an  octahedron  If"  on  a  side.  One  face 
of  the  soHd  is  parallel  to  H  and  5  above  it.  The  horizontal  projection 
of  one  axis  of  the  solid  makes  an  angle  of  30°  with  HA.  Draw  the  H 
and  V  projections  and  also  the  projection  on  a  vertical  plane  parallel 
to  the  axis  described  above. 


282  DESCRIPTIVE  GEOMETRY 

EXERCISE   SHEET   XIX  a   (§51) 

Place  the  sheet  vertically  and  with  the  wide  margin  at  the  left. 
Locate  HA  3"  from  the  upper  border  line. 

Draw  the  H  and  V  projections  of  a  solid  bounded  by  4  regular 
hexagons  and  4  equilateral  triangles,  each  side  of  each  face  being  1" 
long.  Place  the  solid  with  a  hexagon  as  its  base,  at  the  center  of  the 
lower  part  of  the  sheet. 

EXERCISE   SHEET   XX  a   (§54) 

Take  P  at  the  left,  20  from  the  border  line,  turned  away  from  the 
objects,  into  V. 

1.  The  points  a  (13,  0,  3),  b  (22,  0,  12),  c  (19,  0,  15),  and  d  (7,  0,  9) 
determine  the  base  of  a  right  quadrilateral  prism,  2|"  high ;  axis 
vertical.  The  points  e  (4,  16,  4),  /  (4,  5,  10),  and  g  (4,  10,  16)  deter- 
mine the  base  of  a  right  triangular  prism  3"  long;  axis  horizontal. 
Find  the  line  of  intersection  of  the  prisms. 

2.  At  the  right  of  the  above  problem,  draw  the  development  of 
each  of  the  prisms  at  one  half  of  the  former  scale.  Show  the  line  of 
intersection  on  the  development. 

EXERCISE   SHEET   XXI  a   (§§55-62) 

Take  P  at  the  left,  20  from  the  border  line,  and  turned  away  from 
the  objects,  into  H. 

1 .  Draw  the  H,  V,  and  P  traces  of  the  plane  R '  (20,  R^  30°  I,  R'^  45°  0 . 

2.  Given  the  points  a  (II,  30,  3,  17),  b  (II,  52,  3,  17),  c  (II,  30,  11,  7), 
and  d  (II,  52,  11,  7).  Find  the  H,  V,  and  P  traces  of  the  plane  S  con- 
taining the  lines  ab  and  cd. 

EXERCISE   SHEET   XXII  a    (§64) 

[Note.  Unless  otherwise  indicated,  hereafter  P  will  be  taken  at 
the  right  border  line  and  no  profile  projections  will  be  required.] 

1.  Given  the  plane  R  '  (75,  R^  90°,  R''  30°  r) ;  on  R  draw  a  line  ab 
parallel  to  and  6  from  V,  and  draw  a  line  cd  parallel  to  and  6  from  H. 

2.  Given  the  plane  S  (48,  S^  45°  r,  .S"  60°  r) ;  on  S  locate  a  point 
e  at  35  from  P  and  8  from  V. 

3.  Given  the  points  /  (17,  3,  13)  and  g  (7,  7,  3) ;  through  fg  pass 
three  planes  T,  U,  and  W,  the  plane  T  being  perpendicular  to  H. 

1  For  explanation  of  the  n(jtation  used  in  describing  planes,  see  descrip- 
tion of  quantities,  page  xiii. 


APPENDIX  283 

EXERCISE   SHEET   XXIII  a    (§  C6) 

1.  Given  the  points  a  (III,  74,  8,  10)  and  b  (58,  3,  13) ;  through  ab 
pass  the  plane  X,  intersecting  HA  at  60. 

2.  Given  the  points  c  (44,  2,  8)  and  d  (34,  11,  8) ;  through  cd  pass 
the  plane  Y,  intersecting  HA  at  50. 

3.  Given  the  point  e  (16,  5,  9) ;  through  e  pass  the  plane  Z,  inter- 
secting HA  at  21. 

EXERCISE   SHEET   XXIV  a   (§68) 

1.  Through  the  point  a  (68,  3,  7)  pass  two  lines.  The  H  projection 
of  one  line  makes  an  angle  of  60°  to  the  left  and  the  V  projection  an 
angle  of  30°  to  the  right  with  HA.  The  H  and  V  projections  of  the 
other  line  make  angles  of  45°  to  the  right  and  60°  to  the  left,  respec- 
tively.    Draw  the  traces  of  the  plane  X  determined  by  these  lines. 

2.  Given  the  line  joining  the  points  b  (45,  15,  3)  and  c  (33,  15,  15) 
and  the  point  d  (40,  6,  8).  Draw  the  traces  of  the  plane  Y  containing 
the  point  d  and  the  line  be. 

3.  Given  the  triangle  whose  vertices  are/  (18,  Q,2),  g  (10,  12.  5),  and 
h  (6,  4,  13) ;  find  the  traces  of  the  plane  Z  which  contains  the  triangle. 

EXERCISE   SHEET   XXV  a    (§71) 

1.  Given  the  plane  M  (58,  M'' 30°  Z,  M"  4:5°  I)  and  the  point  a 
(III,  69,  13,  5) ;  through  a  draw  a  line  parallel  to  M. 

2.  Given  the  plane  N  (35,  .V^  45°  /,  N^  75°  r)  and  the  point  b  (II, 
46,  4,  10) ;   through  b  draw  a  line  parallel  to  A''. 

3.  Given  the  points  c  (III,  25,  5,  25),  d  (III,  17,  20,  6),  e  (15,  5,  3), 
and  /  (5,  11,  13) ;  draw  the  plane  O  through  the  line  ef  parallel  to  the 
line  cd. 

EXERCISE   SHEET   XXVI  a    (§72) 

1.  Given  the  points  g  (78,  18,  15),  h  (70,  18,  6),  j  (72,  13,  2),  and 
k  (64,  3,  8) ;  through  the  line  jk  pass  the  plane  Q  parallel  to  the  line  gh. 

2.  Given  the  points  I  (51,  14,  4),  m  (42,  3,  8),  and  n  (38,  9,  6); 
through  n  pass  the  plane  R  parallel  to  the  line  Im. 

3.  Given  the  points  ??  (Ill,  24,  9,  7),  q  (III,  16,  3,  7),  and  o  (0,  6, 
7) ;  through  o  pass  the  plane  S  parallel  to  the  line  pq. 


284  DESCRIPTrV^E  GEOMETRY 

EXERCISE   SHEET   XXVII  a    (§74) 

1.  Given  the  plane  T  {57,  T^^  45°  /,  T'  30°  I)  and  the  point  a  (71, 
5,  7) ;  through  a  pass  the  plane  U  parallel  to  T.  (Use  a  normal  case 
auxiliary  line.) 

2.  Given  the  plane  W  (50,  W^  ^o°  r,  TT^'-' 60°  r)  and  the  point  6 
(III,  43,  5,  9) ;  through  h  pass  the  plane  X  parallel  to  W.  (Use  a 
special  case  auxiliary  line.) 

3.  Given  the  plane  Y  (4,  Y^  75°  r,  F"  45°  0  and  the  point  c  (IV, 
13,  11,  5) ;   through  c  pass  the  plane  Z  parallel  to  Y. 

EXERCISE    SHEET    XXVIII  a    (§75) 

Find  the  line  of  intersection  of  each  of  the  following  pairs  of  planes. 

1.  G  (74,  G''  30°  r,  G"  75°  r)  and  H  (56,  m  60°  I,  H^'  45°  /). 

2.  I  (49,  I''  90°,  I^  45°  r)  and  K  (30,  K"^  45°  I,  K^  60°  /)• 

3.  L  (22,  L''  75°  r,  L"  60°  r)  and  M  (10,  JP  30°  r,  M^'  45°  Z). 

EXERCISE    SHEET    XXIX  a    (§§  76,  77) 

Find  the  line  of  intersection  of  each  of  the  following  pairs  of  planes. 

1.  N  (59,  iV*  15°  I,  iV"  60°  r)  and  0,  parallel  to  HA  ;  O''  is  9  below 
and  0""  is  12  above  HA. 

2.  P  (49,  P'^  60°  r,  P"  82^°  r)  and  Q  (28,  Q^  90°,  Q^'  75°  Z). 

3.  i?,  parallel  to  HA  ;  J?''  is  14  below  and  R"  is  17  above  HA ;  and 
S,  parallel  to  HA  ;  S^  is  27  above  and  *S^'  is  4  above  HA. 

EXERCISE    SHEET    XXX  a    (§§75-77) 

Find  the  line  of  intersection  of  each  of  the  following  pairs  of  planes. 

1.  T  (74,  T"^  60°  r,  T^'  45°  r)  and  L^  (64,  U^  60°  r,  C/^' 75°  r). 

2.  X  (32,  X''60°Z,  .Y"30°0  and  W,  parallel  to  i/A.  W  is  11 
below  and  TF"  is  6  above  //A. 

3.  Y  (23,  y"  45°  r,  7"  75°  r)  and  Z  (5,  Z^  60°  Z,  Z"  90°). 

EXERCISE   SHEET   XXXI  a    (§78) 

1.  Find  the  trace  of  the  line  joining  the  points  a  (74,  7,  10)  and  h 
(III,  57,  9,  3)  on  the  plane  Z  (77,  Z''  45°  r,  Z^'  60°  r).  Use  a  normal-case 
auxiliary  plane. 

2.  Find  the  trace  of  the  line  joining  the  points  c  (47,  7,  10)  and 
d  (III,  30,  9,  3)  on  the  plane  Y  (50,  Y^  45°  r,  7"  60°  r).  Use  a  special- 
case  auxiliary  plane.     Compare  result  with  that  of  Ex.  1. 

3.  Find  the  trace  of  the  line  joining  the  points  c  (18,  8,  2)  and 
/  (7,  8,  6)  on  the  plane  X  (10,  X'^  30°  Z,  X-^^  60°  r). 


APPENDIX  285 

EXERCISE   SHEET   XXXII  a   (§78) 

Take  P  at  the  right  border  line,  turned  toward  the  objects,  into  V. 
Use  P  only  where  necessary. 

1.  Find  the  trace  of  the  line  joining  the  points  g  (75,  9,  12)  and 
h  (62,  15,  19)  on  the  plane  W  (59,  W"  90°,  T7^  60°  I). 

2.  Find  the  trace  of  the  line  joining  the  points  i  (III,  48,  15,  14) 
and  J  (IV,  35,  4,  8)  on  the  plane  U,  which  is  parallel  to  HA,  JJ^  being 
12  below,  and  f/^'  9  above  HA . 

3.  Find  the  trace  of  the  line  joining  the  points  k  (20,  8,  5)  and 
I  (IV,  10,  4,  1)  on  the  plane  T,  which  passes  from  I  to  III  throuTh  HA, 
and  makes  an  angle  of  30°  with  H. 

EXERCISE   SHEET   XXXIII  a    (§§79-82) 

1.  On  the  plane  M  (60,  M^  45°  I,  M^  60°  I),  locate  a  point  a  72  from 
P  and  6  from  V.     Draw  the  line  ab  1^"  long  and  perpendicular  to  M. 

2.  Given  the  points  c  (45,  6,  3)  and  d  (32,  13,  10) ;  through  c  pass 
a  plane  A^  perpendicular  to  cd. 

3.  Given  the  points  e  (II,  25,  5,  13),  /  (II,  16,  11,  9),  and  g  (IV, 
10,  7,  15) ;  through  g  pass  a  plane  0  perpendicular  to  ef. 

EXERCISE    SHEET    XXXIV  a    (§85) 

1.  Draw  R  parallel  to  (?  (61,  Q''  45°  r,  Q^  60°  r),  and  I"  from  Q. 

2.  Given  the  points  h  (III,  52,  2,  3),  i  (30,  12,  13),  andj  (43,  11,  9) ; 
find  the  shortest  distance  from  j  to  the  line  hi. 

3.  Given  the  plane  T  (23,  T^  30°  r,  T^  45°  r)  and  the  point  m  (13, 
6,  10) ;   find  the  shortest  distance  between  m  and  T. 

EXERCISE    SHEET    XXXV  a    (§87) 

1.  Given  the  points  a  (65,  15,  2),  6  (IV,  54,  5,  12),  c  (55,  1,  17),  and 
d  (40,  7,  2) ;  find  the  projections  and  true  length  of  the  shortest  line 
between  the  lines  ab  and  cd. 

2.  Given  the  points  a  (21,  18,  16),  b  (8,  5,  2),  and  c  (III,  11,  2,  13). 
Through  c  draw  the  line  cd,  parallel  to  ab,  and  find  the  common  per- 
pendicular between  ab  and  cd. 

EXERCISE  SHEET  XXXVI  a  (§90) 
Draw  the  line  of  greatest  declivity  of  each  of  the  following  planes 
with  respect  to  H,  and  also  with  respect  to  V.  Determine  the  true 
angle  that  each  plane  makes  with  H,  and  also  with  V.  The  planes 
are  X  (60,  X^  45°  I,  X^  60°  I) ;  V  (50,  V^  45°  I,  V-  30°  r)  and  Z,  parallel 
to  HA.    ZZ^  is  8  below  and  ZZ^  20  above  HA. 


286  DESCRIPTIVE  GEOMETRY 

EXERCISE   SHEET   XXXVII  a    (§91) 

1.  Given  the  point  a  at  distances  71  from  P  and  10  from  H,  and 
the  point  6  at  distances  75  from  P  and  7  from  V ;  and  given  that  both 
a  and  b  lie  in  X  (60,  X''  45°  I,  X"  60°  I).     Rabatte  a  into  H  and  b  into  V. 

2.  Given  the  point  c  at  distances  39  from  P  and  10  from  H  and 
lying  in  the  plane  Y  (44,  Y^  45°  r,  F^'  75°  I).  Rabatte  c  into  H.  Also 
rabatte  FF"  into  H. 

3.  The  four  points  e,  f,  g,  and  A  all  lie  in  Z  (14.  Z''  30°  Z,  Z"^  45°  0  ; 
e  is  in  I  23  from  P  and  6  from  H,  rabatte  e  into  V ;  /is  in  III,  3  from 
P  and  6  from  H,  rabatte/  into  H ]  (7  is  in  II,  15  from  P  and  4  from  F, 
rabatte  g  into  F;  /^  is  in  IV,  12  from  P  and  9  from  V,  rabatte  h  into  H. 

EXERCISE   SHEET   XXXVIII  a    (§§93,94) 

1.  The  four  points  a,  b,  c,  and  d  he  on  the  plane  S  (46,  5^^  30°  I, 
S"  45°  I)  ;  a  is  70  from  P  and  12  from  H  ;  6  is  62  from  P  and  11  from  H; 
c  is  55  from  P  and  5  from  H ;  c?  is  60  from  P  and  3  from  H.  Find  true 
shape  of  quadrilateral  determined  by  these  points  by  rabatting  into  H. 

2.  Given  the  points  e  (30,  3,  7),  /  (17,  12,  8),  and  g  (21,  3,  4).  Find 
the  true  angle  efg  by  rabattement  into  V. 

EXERCISE   SHEET   XXXIX  a    (§96) 

1.  Given  the  plane  TF  (72,  TF^' 30°  r,  TF'' 45°  r) ;  draw  the  projec- 
tions of  a  six-pointed  star,  inscribed  in  a  2"  circle  and  lying  in  TF. 

2.  Given  the  plane  X  (19,  V' 60°  r,  V^' 75°  0  ;  draw  the  projec- 
tions of  an  equilateral  triangle  1|"  on  a  side  lying  in  X  with  one  side 
parallel  to  F. 

EXERCISE    SHEET   XL  a    (§97) 

Given  the  plane  Z  (59,  Z^  45°  r,  Z"  30°  r),  and  using  a  scale  of  f "  = 
1",  draw  the  projections  of  a  hollow  brick,  2"  X  4"  X  8",  resting  with 
one  of  its  2"  X  4"  faces  in  contact  with  Z.  Let  one  of  the  4"  edges  of 
this  face  make  an  angle  of  15°  with  ZZ''.  Keep  the  entire  soUd  within 
the  first  quadrant. 

EXERCISE   SHEET   XLI  a    (§98) 

1.  Given  the  plane  X  (75,  X''  30°  r,  X"  45°  r^)  and  the  points  e  (71, 
13,  7)  and  /  (57,  5,  7) ;   find  the  angle  between  ef  and  X. 

2.  Given  the  plane  \V  (21,  IF''  45°  r,  IF"  45°  I)  and  the  points  g  {31, 
0,  15)  and  h  (18,  13,  8) ;  find  the  angle  between  gh  and  W. 


APPENDIX  287 

EXERCISE   SHEET   XLII  a   (§99) 

1.  The  plane  Z  cuts  HA  at  the  point  (77,  0,  0).  The  trace  ZZ^ 
makes  an  angle  of  60°  r  with  HA  and  Z  makes  an  angle  of  75°  with  Y . 
Find  ZZ^. 

2.  The  plane  Y  cuts  HA  at  the  point  (50,  0,  0).  The  trace  FF^ 
makes  an  angle  of  30°  r  with  HA  and  Y  makes  an  angle  of  30°  with  H. 
Find  YY^. 

3.  The  plane  X  cuts  HA  at  the  point  (3,  0,  0).  The  trace  XX^ 
makes  an  angle  of  60°  I  with  HA  and  X  makes  an  angle  of  45°  with  HA. 
Find  XXk 

EXERCISE   SHEET   XLIII  a    (§100) 

Using  auxiliary  spheres  \\"  in  diameter  with  their  centers  at  the 
points  (i)  (65,  0,  0),  {2)  (40,  0,  0),  and  (3)  (14,  0,  0),  respectively,  draw 
the  traces  of  the  planes  that  make  the  following  angles  with  H  and  7. 

1.  The  plane  R\  45°  with  Y  and  60°  with  H. 

2.  The  plane  S ;   30°  with  7  and  60°  with  H. 

3.  The  plane  T ;  30°  with  H  and  30°  with  7. 

EXERCISE   SHEET   XLIV  a    (§100) 

[Note.  When  the  first  plane  is  found,  use  an  auxiliary  cone  with 
its  apex  at  a.] 

1.  Through  the  point  a  (60,  5,  10),  pass  as  many  planes  as  possible 
that  make  angles  of  45°  with  7  and  60°  with  H,  respectively. 

2.  Given  the  points  c  (23,  2,  13)  and  d  (16,  8,  3) ;  through  cd  pass 
as  many  planes  as  possible,  making  angles  of  45°  with  7.  Is  there  an 
impossible  case  for  this  problem  ? 

EXERCISE   SHEET   XLV  a    (§101) 

Find  the  angle  between  each  of  the  following  pairs  of  planes. 

1.  Z  (75,  Z^  60°  r,  Z-  30°  r)  and  Y  (55,  F'^  30°  Z,  F-  60°  l). 

2.  X  (50,  X'^  60°  r,  Z-  30°  r)  and  W  (35,  W'  60°  r,  W-  45°  I). 

3.  V   (25,    f/*60°r,    C7M5°  r)   and   7   (5,   7^90°,   7''45°Z).     (See 

Art.  76.) 

EXERCISE   SHEET   XL VI  a    (§101) 

Find  the  angle  between  each  of  the  following  pairs  of  planes. 

1.  0  (75,  0^  15°  r,  O"  90°)  and  P  (55,  P^  90°,  P"  30°  I). 

2.  Q  (46,  Q'^  60°  I,  Q'  60°  Z)  and  R  (34,  P'"*  45°  r,  P"  45°  r). 

3.  S  (S''  is  parallel  to  and  12  above  HA ;  iS^  is  12  below  HA)  and 
T  (15,  T^  45°  r,  T"  60°  r). 


288 


DESCRIPTIVE  GEOMETRY 


EXERCISE   SHEET   XLVII  a    (§§102,103) 

1.  The  point  a  (68,  3,  11)  is  the  center  of  the  base  of  a  cube  whose 
edge  is  I4"  long.  The  base  is  parallel  to  H  and  one  of  its  body  diagonals 
is  parallel  to  V.  Find  the  points  where  the  line  joining  the  points 
h  (77,  1,  8)  and  c  (58,  13,  17)  pierces  this  cube.     Mark  them  d  and  e. 

2.  The  points  /  (47,  0,  2),  g  (47,  0,  14),  and  h  (37,  0,  8)  define  the 
base  of  a  regular  tetrahedron  (§50).  Find  the  points  k  and  I  in  which 
the  line  joining  the  points  i  (50,  0,  15)  and  j  (36,  10,  2)  pierces  the 
tetrahedron. 

3.  The  points  m  (15,  0,  12),  n  (8,  0,  12),  o  (13,  0,  18),  and  p  (6,  0, 
18)  define  the  base  of  a  quadrilateral  prism.  The  points  m  and  q 
(21,  11,  2)  determine  one  edge  of  the  prism.  Find  the  line  of  inter- 
section of  the  plane  Y  (29,  Y^  45°  r,  7"  45°  r)  and  the  given  prism. 

EXERCISE   SHEET   XL VIII  a    (§104) 

Place  the  sheet  vertically  with  wide  margin  at  the  left.  Given  the 
tetrahedron  a  (46,  6,  4),  h  (8,  2,  21),  c  (14,  25,  28),  d  (25,  27,  39),  and 
the  pyramid  e  (31,  29,  19),  /  (46,  0,  22),  g  (41,  0,  4),  h  (16,  0,  16),  i 
(21,  0,  32).     Find  the  Hne  of  intersection  of  the  soHds. 

__  [Note.     Exercise  sheets  XLIX  a  to 

LVIII  a  are  given  as  applications  of  the 
principles  of  Chapter  VII,  but  without 
special  reference  to  specific  articles.  Each 
exercise  is  intended  to  be  solved  on  a 
standard  sheet,  8"  X  10^".  The  exact 
layout  for  each  sheet  is  to  be  determined 
by  the  student.] 


a 

« 

b 

?♦-- 

-fl 

'    1 

.     .      1^2." 

ii 

1 

^\-^ 

,\ 

Fig.  255 

Cv 

EXERCISE   SHEET   XLIX  a 

The  garage  door  shown  in  Fig.  255  is  hung  from  two  trolleys,  pivoted 
at  a  and  h.  The  trolley  a  moves  on  the 
track  ab,  while  h  moves  on  6c,  as  shown 
by  the  arrows.  Determine  the  clear- 
ances needed  for  the  operation  of  the 
door.     Scale,  l"  =  l'-0". 


Fixed  pin 


EXERCISE    SHEET   XLIX  b 

Figure  256  shows  the  hinged  front 
of  a  writing  desk,  supported  when  open 
by  a  slotted  metal  rod  at  the  side.     De- 


3.*-- 


\'Jy/}^////////M 


Fig.  256 


APPENDIX  289 

termine  the  path  of  travel  traced  by  the  end  of  the  rod,  as  the  desk 
closes.  Also  show  the  entire  area  over  which  any  part  of  the  rod 
moves.     Scale,  3"  =  1'  —  0". 

EXERCISE    SHEET   La 

Figure  257  shows  the  path  followed  by  the  front  wheel  of  a  motor- 
cycle equipped  with  a  side  car.  Let  A  =  4'  —  6";  B  =  2'  —  6"; 
C  =  3'  -  6".     Let  both  wheels  of  the 

motorcycle  follow  the  same  path.    Trace  [ 

the  wheel  track  left  by  the  side  car.  a 

Scale,  h"  =  1'  -  0".      '  ,^'- :<^i^-L 


EXERCISE    SHEET   LI  a 


^S 


Figure  257  shows  the  path  followed 
by   the  front  wheel  of    a    motorcj'cle  "         Fiq,  257 

equipped    with   a   side    car.     Let   A  = 

4'  -  0";  5  =  3'-  0";  C  =  4'  -  6".  Let  it  be  assumed  that  the 
front  wheel  moves  at  a  uniform  velocity  and  that  the  rear  wheel  moves 
as  slowly  as  is  consistent  with  the  motion  of  the  front  wheel.  Trace 
the  path  of  both  wheels  of  the  •motorc3'cle  and  of  the  wheel  of  the 
side  car.     Scale,  h"  =  V  -  0". 

EXERCISE    SHEET   LII  a 

Draw  the  front  elevation  of  an  elUptical  arch  with  a  span  of  24'  —  0" 
and  a  rise  of  8'  —  0".  Scale,  j"  =  V  —  0".  Use  the  approximate 
construction  of  §  127,  IIL  Let  the  arch  consist  of  19  stones,  of  equal 
width  on  the  intrados,  and  of  equal  depth.  Joint  lines  between  stones 
are  made  normal  to  the  curve  of  the  arch.  Let  the  wall  be  2'  —  0" 
thick.  Draw  also  a  plan  of  the  arch  looking  up,  and  a  development  of 
the  soffit.  On  the  elevation,  for  purposes  of  comparison,  trace  the 
outline  of  the  extrados  as  given  by  the  methods  of  §  127, 1  and  §  127,  II, 
as  well  as  that  of  §  127,  III. 

EXERCISE   SHEET   LII  b 

Place  the  sheet  vertically.  HA,  3"  from  the  top  border.  Dr.iw  the 
curve  traced  by  a  point  on  the  circumference  of  the  base  of  a  right 
circular  cone,  which  is  so  placed  that  an  element  is  in  contact  with  H. 
Let  the  cone  roll  on  H,  keeping  the  apex  in  a  fixed  position.  Let  the 
diameter  of  the  cone  be  2j"  and  the  altitude  be  2f ". 


290 


DESCRIPTIVE  GEOMETRY 


EXERCISE   SHEET  LIII  a 

•  Draw  a  set  of  pivoted  links  similar  to  Fig.  142,  making  CD  =  2" ; 
AB  =  U";  CA  =  4",  and  DB  =  something  between  2\"  and  ^". 
Trace  the  path  of  a  point  on  DB,  1"  from  D;  also  of  a  point  on  DB, 
extended  1"  beyond  B. 


EXERCISE   SHEET   LIV  a 

Let  a  circle,  2"  in  diameter,  roll  on  a  straight  line.     Trace  the  path 
of  a  point  which  lies  on  a  radius  of  the  circle,  1"  outside  the  circum- 
ference, and  which  moves  with  the  circle. 
A! 

EXERCISE   SHEET  LV  a 

The  window  sash  shown  in  Fig.  258  swings  on 
the  hinges  A  A.  Determine  how  the  back  of  the 
sash  must  be  cut  to  allow  it  to  swing  open.  Scale, 
one  half  full  size. 

EXERCISE    SHEET   LVI  a 

Place  the  sheet  vertically.     In  the   center  of 
the  sheet  draw  a  circle,  3"  in  diameter,  and  draw 
Fig.  258  its  vertical  axis  AB.     Trace  each  of  the  following 

curves. 

(1)  The  curve  generated  by  a  point  so  moving  that  its  distance 
from  AB  i^  constantly  equal  to  its  distance  from  the  circle. 

(2)  When  its  distance  from  AB  is  constantly  equal  to  twice  its 
distance  from  the  circle. 

(3)  When  its  distance  from  a  border  line  is  constantly  equal  to  its 
distance  from  the  circle. 

(4)  When  its  distance  from  a  corner  of  the  sheet  is  constantly  equal 
to  its  distance  from  the  circle. 


EXERCISE   SHEET  LVII  a 

The  points  a  (38,  10,  2),  6  (12,  24,  9),  c  (8,  2,  23)  determine  the 
plane  of  a  parabola  which  has  its  vertex  at  a  and  which  also  passes 
through  b.     Draw  the  projections  of  the  curve. 


APPENDIX  291 

EXERCISE   SHEET   LVIII  a 

In  the  center  of  the  sheet,  and  with  their  axes  placed  vertically, 
draw  two  intersecting  circular  arcs  each  having  a  1^"  radius,  and  with 
their  centers  2j"  apart.  Draw  the  involute  of  the  curve  thus  estab- 
lished. 

EXERCISE   SHEET   LIX  a    (§§145-148) 

(1)  The  line  a  (54,  15,  15),  h  (61,  6,  5)  is  the  right-hand  terminating 
element  of  an  open  cylinder,  similar  to  the  one  in  Fig.  155.  Draw 
the  projections  of  the  cylinder.  Indicate  at  least  six  elements.  Let 
all  invisible  lines  be  dotted. 

(2)  Given  the  plane  Z  (38,  Z^  45°  I,  Z''45°r).  Draw  the  projec- 
tions of  a  right  circular  cone,  Ij"  in  diameter  and  1^"  high,  which  has 
its  base  in  contact  with  Z.     Let  the  cone  point  downward. 

(3)  The  point  e  (12,  0,  12)  is  the  center  of  the  lower  base  of  a  double 
right  circular  cone,  the  base  of  which  is  2"  in  diameter  and  whose 
elements  make  angles  of  45°  with  H.  Locate  the  following  points  on 
the  surface  of  the  cone ;  /,  16  from  P  and  4  from  H;  g,  S  from  P  and 
12  from  H;  h,  11  from  P,  9  from  H,  and  17  from  V. 

EXERCISE   SHEET   LX  a    (§§  149,  150) 

1.  The  point  a  (68,  0,  13)  is  the  center  of  a  free-hand  spiral  curve 
similar  to  that  in  Fig.  146.  Draw  about  two  turns  within  a  space 
about  two  inches  square.  Let  this  curve  be  a  right  section  of  a  cylinder. 
Find  the  line  of  intersection  between  this  cylinder  and  the  plane  Z 
(69,  Z''60°Z,  ZM5°r). 

2.  The  point  b  (33,  0,  14)  is  the  center  of  a  2"  circle  in  H.  The 
point  c  (10,  24,  4)  is  the  apex  of  an  elliptical  cone  having  the  above 
circle  for  its  base.     Determine  the  right  section  of  the  cone. 

EXERCISE   SHEET  LXI  a   (§151) 

1.  The  point  a  (70,  16,  16)  is  the  center  of  a  circle  2"  in  diameter 
and  which  has  its  base  in  a  plane  perpendicular  to  HA.  This  circle 
is  the  base  of  a  cone  whose  apex  is  at  b  (46,  2,  3).  The  line  m  (69,  4, 
25),  n  (52,  11,  2)  pierces  the  cone.  Find  the  piercing  points.  Note: 
In  this  problem  a  profile  view  of  the  cone  base  is  useful  but  no  space 
has  been  provided  for  it  on  the  sheet.  Let  the  base  circle  be  conceived 
as  being  rotated  on  its  vertical  axis,  and  let  it  be  so  drawn,  using  a'" 
as  a  center.  If  indicated  by  a  light  line,  this  projection  can  be  used 
as  needed  without  confusing  the  drawing. 


292  DESCRIPTn^  GEOMETRY 

2.  The  points  o  (28,  0,  IG),  p  (14,  22,  lOj  are  the  termini  of  the 
axis  of  a  cylinder  whose  base  is  a  2\"  circle  in  H.  The  line  q  (29j  3, 
10),  r  (9,  19,  17)  pierces  the  cj-linder.     Determine  the  piercing  points. 

EXERCISE    SHEET   LXII  a    (§152) 

1.  The  point  a  (61,  0,  14,)  is  the  center  of  the  base  of  a  60°  right 
circular  cone,  3"  in  diameter.  A  right  circular  cylinder,  \\"  in  diameter 
and  2\"  high,  rests  on  H  with  the  center  of  its  base  at  h  (56,  0,  18). 
Find  the  line  of  intersection. 

2.  The  point  c  (24,  0,  14)  is  the  center  of  the  base  of  a  cone  like  the 
one  in  (1)  above.  The  point  d  (36,  0,  22 j  is  the  center  of  a  l\"  circle 
in  H,  which  is  the  base  of  a  cylinder  whose  axis  passes  from  d  to  e  (12, 
16,  9).     Find  the  line  of  intersection. 

EXERCISE    SHEET   LXU  h    (§  152) 

[Note,  All  four  cones  on  this  sheet  are  60°,  right  circular  cones 
with  bases  2\"  in  diameter.] 

1.  Find  the  line  of  intersection  between  a  cone  with  its  apex  at 
a  (20,  12,  0)  and  another  with  the  center  of  its  base  at  h  (17,  15,  0). 
The  axis  of  each  cone  is  horizontal,  and  perpenchcular  to  T'. 

2.  Find  the  line  of  intersection  between  a  cone  having  the  center 
of  its  base  at  c  (69,  0,  12)  and  its  axis  vertical,  with  another  cone  which 
is  tangent  to  H  along  the  hne  d  (76,  0,  12),  e  (56,  0,  12). 

EXERCISE    SHEET   LXUI  a    (§152) 

1.  The  point  a  (66,  0,  11  j  is  the  center  of  the  base  of  a  2\",  60°,  right 
circular  cone.  The  line  c  (.58,  10,  16),  d  (73,  10,  7)  is  the  axis  of  a  2n" 
right  circular  cone.  Draw  the  Une  of  intersection.  It  will  be  found 
convenient  to  use  a  V  plane  passed  parallel  to  the  base  of  the  latter 
cone. 

2.  The  point  e  (26,  0.  10)  is  the  center  of  the  base  of  a  right  circular 
cone,  2f"  high  and  2^"  in  diameter.  A  right  circular  cylinder,  If"  in 
diameter,  has  its  axis  in  the  line/  (21,  7,  18),  g  (34,  7,  5).  Find  the 
line  of  intersection. 

EXERCISE   SHEET  LXIV  a    (§152) 

The  line  a  (15.  8.  16),  6  (32,  26.  16)  is  the  axis  of  a  3"  right  circular 
cone.  The  point  c  (28,  0,  16)  is  the  center  of  the  base  of  a  right  cir- 
cular cone,  3"  in  diameter  and  4:\"  high.  Find  the  hne  of  intersection 
by  the  concentric  spheres  method.     Place  the  sheet  vertically. 


APPENDIX  293 

EXERCISE   SHEET  LXV  a   (§§  153,  154) 

1.  The  line  a  (69,  4,  10),  6  (62,  17,  23)  is  the  axis  of  a  cyHnder  whose 
base  is  a  1"  circle  parallel  to  H.  Determine  a  point  c,  on  the  cylinder, 
8  from  H  and  16  from  V.  Through  c  draw  a  plane  Z,  tangent  to  the 
cylinder. 

2.  The  point  d  (34,  0,  10)  is  the  apex  of  a  60°  right  circular  cone 
whose  axis  is  perpendicular  to  H.  Through  e  (39,  4,  5)  draw  the 
planes  X  and  Y,  tangent  to  the  cone.  (An  auxiliary  H  plane  passed 
through  the  base  of  the  cone  will  be  found  useful.) 

3.  The  line/  (14,  7,  0),  q  (4,  4,  16)  is  the  axis  of  a  cone  whose  base 
is  a  IV'  circle  in  V.  Through  h  (2,  18,  7)  draw  the  planes  V  and  TF, 
tangent  to  the  cone.  (In  determining  the  traces,  do  not  overlap  prob- 
lem 2.) 

EXERCISE   SHEET   LXVI  a    (§§  155,  156) 

Draw  a  cjdinder  and  two  cones  like  those  for  sheet  LXV  a. 

1.  Pass  a  plane  Z,  tangent  to  the  cylinder  and  parallel  to  the  line 
a  (61,  10,  4),  6  (53,  4,  5). 

2.  Through  the  point  c  (44,  7,  20) ;  draw  three  Hnes  cd,  ce,  and  c/, 
each  tangent  to  the  cone. 

3.  Through  g,  which  lies  on  the  cone,  8  from  H  and  6  from  V , 
draw  a  line  hj,  tangent  to  the  cone. 

EXERCISE   SHEET  LXVH  a   (§157) 

In  the  upper  left-hand  corner  of  the  sheet,  draw  a  right  circular 
cyhnder,  1"  in  diameter  and  2"  long,  in  a  vertical  position.  Let  this 
cyHnder  be  intersected  by  an  equal  cylinder  in  a  horizontal  position; 
and  let  this  latter  be  intersected  by  still  another  which  is  inclined  at 
45°  to  H.  Let  the  length  of  all  three  cylinders  be  equal  and  let  all 
the  axes  intersect.  Draw  the  development  of  each.  On  the  develop- 
ments, and  also  on  the  elevation,  draw  a  helix,  inclined  at  30°  to  the 
axes  of  the  cylinders,  and  which  is  continuous  on  all  three  parts. 

EXERCISE   SHEET  LXVIH  a    (§§158-160) 

Draw  the  projections  of  a  developable  helicoid  of  two  nappes,  based 
on  a  cylinder  \\"  in  diameter  and  3"  high.  Place  the  sheet  vertically 
with  HA  in  the  center.  Locate  the  center  of  the  base  of  the  cylinder 
at  a  (28,  0,  28).  Draw  carefully  to  bring  out  the  separate  nappes,  and 
to  show  the  visible  and  invisible  portions. 


294  DESCRIPTIVE  GEOMETRY 

EXERCISE   SHEET   LXIX  a    (§§163,164) 

Draw  the  projections  of  a  If"  cube  placed  with  one  body  diagonal, 
parallel  to  HA.  Rotate  the  cube  on  this  diagonal  and  draw  the  pro- 
jections of  the  surface  that  results.  Before  concluding  the  exercise, 
the  student  should  account  for  the  surface  generated  bj'-  each  of  the 
12  edges. 

EXERCISE    SHEET   LXIX  b    (§§163,164) 

Repeat  sheet  LXIX  a,  using  a  right  parallelepiped,  1"  X  2"  X  3". 

EXERCISE   SHEET  LXX  a   (§165) 

Place  the  sheet  vertically  and  divide  into  four  equal  parts.  The 
larger  sheet  will  give  better  results.  Draw  the  plan  and  elevation  of 
four  different  warped  surfaces,  including  at  least  one  whose  generation 
is  dependent  on  a  time  factor.  Show  the  directrices  of  each  surface. 
Let  at  least  one  of  the  surfaces  be  cut  by  a  plane  perpendicular  to  H 
or  V  and  trace  the  line  of  intersection. 

EXERCISE   SHEET   LXXI  a    (§§  164,  165) 

Place  the  sheet  vertically  with  HA  in  the  center.  Draw  the  line 
a  (28,  0,  20),  6  (28,  36,  20).  Let  ah  be  the  axis  of  a  right  heHcoid  (§  165), 
making  two  turns  in  the  height.  Let  the  lines  c  (16,  0,  38),  d  (50,  36, 
38),  and  e  (9,  36,  2),  /  (46,  0,  2)  and  the  plane  H  be  the  directrices  of 
an  hyperboHc  paraboloid.     Find  the  line  of  intersection  of  the  surfaces. 

EXERCISE   SHEET  LXXII  a    (§§166-172) 

Place  the  sheet  vertically.  Take  HA  in  the  center.  Draw  the 
projections  of  a  torus  generated  by  rotating  a  1|"  sphere  about  a 
vertical  axis  passing  through  m  (36,  10,  20).  Let  the  center  of  the 
sphere  be  l\"  from  the  axis.  Locate  the  points  o  and  o',  on  the  torus, 
24  from  P  and  13  from  V.  Find  the  traces  of  the  planes  Z  and  Z' 
which  are  tangent  to  the  torus  at  o  and  o'.  Draw  two  lines  tangent 
to  the  torus  and  one  which  pierces  the  torus  at  o  and  at  another  point, 
q,  to  be  determined. 

EXERCISE   SHEET  LXXIII  a   (§173) 

Place  the  sheet  vertically,  HA  3|"  from  the  top.  Let  a  (34,  11,  11) 
be  the  center  of  a  2|"  sphere  and  b  (46,  25,  24)  be  the  apex  of  a  cone 
tangent  to  the  sphere.  Draw  the  projections.  Extreme  accuracy  of 
construction  is  necessary  if  the  points  of  tangency  are  to  check  properly 
on  cone  and  sphere. 


APPENDIX  295 

EXERCISE    SHEET   LXXIV  a    (§§175-177) 

Place  the  sheet  vertically,  HA  in  the  center.  Let  a  (25,  0,  18)  be 
the  center  of  the  base  of  a  45°  right  circular  cone  4"  m  diameter.  Let 
b  (34,  8,  20)  be  the  center  of  an  annular  torus  formed  by  rotating  a 
1|"  sphere  about  a  vertical  axis  through  b,  the  radius  of  rotation  being 
1|"  to  the  center  of  the  sphere.     Find  the  line  of  mtersection. 

EXERCISE    SHEET   LXXIV  b    (§§175-177) 

Make  up  a  sheet  similar  to  Fig.  185,  using  a  vase  form  like  that 
shown  m  Fig.  183,  intersected  by  a  right  circular  cylinder.  Draw 
the  figures  as  large  as  possible. 

EXERCISE   SHEET  LXXV   (§181) 

Construct  two  sections  through  the  building  shown  in  Fig.  7,  at 
about  three  times  the  scale  of  the  figure.  Let  one  section  be  cut  by  a 
plane  perpendicular  to  the  longitudinal  plan  axis  and  the  other  by  a 
plane  perpendicular  to  H  and  at  45°  to  the  same  axis. 

[Note.  Figures  194,  197,  198  may  be  used  for  a  further  study  of 
sections  in  a  manner  similar  to  the  above.] 

EXERCISE    SHEET   LXXVI    (§182) 

Construct  a  plan  and  elevation  of  a  corner  rafter  similar  to  Figs. 
188  and  189,  but  at  three  times  the  scale.  Design  a  different  outline 
for  the  jack  rafter  and  increase  or  decrease  the  pitch  of  the  roof.  Deter- 
mine the  proper  shape  for  the  hip  rafter. 

This  problem  may  be  varied  by  requiring  a  reentrant  corner  to  be 
drawn. 

EXERCISE   SHEET   LXXVII    (§183) 

Construct  a  plan  and  elevation  similar  to  Fig.  191,  but  at  three 
times  the  scale.  Find  the  openings  required  for  a  shaft,  the  right 
section  of  which  is  an  equilateral  triangle,  If"  on  a  side. 

EXERCISE    SHEET   LXXVIII    (§§184-187) 

Use  a  large  sheet,  placed  vertically  and  divided  into  four  parts  as 
shown  on  p.  xii.     Omit  HA  lines. 

Construct  an  elevation  and  a  roof  plan  similar  to  Fig.  197  for  each 
of  the  cases  given  in  the  table  below.  Use  jV"  sscale.  The  dimensions 
in  the  table  are  in  feet  and  are  to  be  used  to  replace  the  letters  in  Fig.  197. 


296 


DESCRIPTIVE  GEOMETRY 


APPENDIX 


207 


Letters  on  Figure  197 
Case      ahcdefgh 

1  40     20     20    40     25     25     10     10 

2  40    20    20    40     12     30      8     12 

3  40     20     20     40     25     10     12      8 

4  Using  dimensions  as  in  2,  above, 
let  the  low  wing  be  revolved  on  0 
until  the  angle  is  60°. 

[XoTE.  Donners  similar  to  those  in  Figs.  192  /  and  198  may  be 
added  if  desired.  Let  all  roofs  have  a  slope  of  45°  and  the  extreme 
overhang  in  anj'  direction  be  2'  —  0".] 

EXERCISE    SHEET   LXXIX    (§§184-187) 

Use  a  large  sheet  placed  horizontally. 

1.  Figures  259,  260  show  a  floor  plan  and  two  elevations  of  a  build- 
ing with  a  somewhat  complicated  roof.  Draw  a  roof  plan  an  place 
of  the  floor  plan  in  the  figure),  four  elevations,  and  two  sections  at 
■g^"  scale.  Sufficient  mformation  is  given  to  make  it  possible  to  con- 
struct the  required  drawings.  Particular  attention  should  be  given 
to  points  A  and  B,  in  working  out  the  roof. 

2.  Cast  sun  shadows  on  the  Dlan  and  front  elevation. 


6i' 


'z. 

N 

W-hD— E 


t-»  


All  eaves    project  i-s' 
All  roofs  e"1hic.k.. 
.^  Pitch  as  shown  below. 


ROOf     PUM. 


'I  -* 


Fig.  260 


298 


DESCRIPTIVE  GEO.METRY 


All  wolls  and  roofs  I'-oMhick. 
All  roofs  slope  A5'  and  pro- 
ject 2'  beyond  walls. 


fT.  is'  above  ground  for 
all  main  roob;  3?' 
for  tower. 


P'iG.    261 


APPENDIX 


299 


EXERCISE   SHEET  LXXX   (§188) 

Use  a  large  sheet,  placed  vertically. 

Figure  261  shows  a  building  with  a  tall  chimney.  Draw  the  plan 
and  elevation  at  I"  scale.  Four  guy  wires  are  to  be  attached  to  the 
stack.  They  shall  be  in  planes  perpendicular  to  H  and  at  45°  with  F, 
and  passing  through  the  center  of  the  chimney.  The  wires  are  to 
make  -angles  of  45°  with  H.  Find  where  the  wires  will  be  attached  to 
the  building  and  the  angles  between  the  wires  and  roof. 

EXERCISE   SHEET   LXXXI    (§189) 

Use  a  large  sheet 
placed  vertically. 

Draw  the  plan,  ele- 
vation, and  section  of 
a  hipped  roof  frame 
similar  to  Fig.  199, 
except  that  the  end 
plane  of  the  roof  shall 
be  inclined  at  45°  to 
H  while  the  side  planes 
remain  at  30°.  Deter- 
mine all  10  angles  as 
described  in  §  189. 
Let  the  total  width  be 
11'  -  0";  rafters  3" 
X  8".     Use  f "  scale. 

EXERCISE    SHEET 
LXXXII    (§  189) 

Use  large  sheet 
placed  vertically. 

Draw  plan,  eleva- 
tion, and  section  and  determine  all  10  angles  as  described  in  §  189  for 
the  hipped  roof  frame  shown  in  Fig.  262.     Use  |"  scale. 

EXERCISE    SHEET   LXXXIII    (§190) 

Draw  the  plan  and  section  of  a  sheet  metal  chute  similar  to  Fig.  204, 
#2,  at  f"  scale.  Draw  the  plan,  elevation,  and  development  of  the 
clip  at  3"  scale  and  determine  the  angle  of  the  bend,  after  the  manner 
of  Figs.  206  and  207.     Use  large  sheet. 


Rafters    is"  c-c 

All  timbers    3"  thick. 

Roof  planer  moke  30°  with  H. 

Fig.  262 


300 


DESCRIPTIVE  GEOMETRY 


EXERCISE    SHEET   LXXXIV    (§§191,    192) 
Use  large  sheet  placed  horizontally  and  divided  vertically  through 
the  center. 

1.  (On  left  half  of  the  sheet.)  Draw  the  plan  and  elevation  of  a 
vase  form  similar  to  Figs.  183,  185,  or  242,  using  as  large  a  scale  as 
convenient.  Assume  the  vase  to  be  turned  from  a  stick  of  wood  with 
the  grain  centered  as  in  Fig.  208.  Use  not  less  than  10  annual  rings. 
Find  the  grain  pattern  on  the  elevation. 

2.  (On  right  half  of  the  sheet.)  Draw  the  plan  and  elevation  of  a 
spiral  chute  like  Fig.  209,  at  ^"  scale.  Let  the  story  height  be  10'  — 
0"  in  the  clear;  floor  thickness  1'  —  0";  diameter  of  center  support 
0'  —  8" ;  diameter  of  outside  of  chute  7'  —  0" ;  vertical  height  of 
edge  0'  —  9".  Develop  outer  and  inner  helixes  at  l"  scale.  Develop 
well  lining  at  ^"  scale.  Draw  plan,  elevation,  and  approximate  de- 
velopment of  Y^j  of  one  turn  of  the  helicoidal  bottom  at  H"  scale. 
Figure  all  details  to  show  accurate  sizes. 

EXERCISE    SHEET   LXXXV    (§193) 

Use  large  sheet.     Figure  263  shows  the  layout. 
1.   Find  the  line  of  intersection  of  both  the  inner  and  outer  surfaces 
of  the  vaults. 


Fig.  263 


APPENDIX 


301 


2.  (a)  The  line  of  intersection  of  two  vaults  is  given ;    also  the 
section  of  the  larger  vault ;    find  the  right  section  of  the  smaller  vault. 

(b)  Assume  an  arbitrary  hne  of  intersection  on  plan  and  find  the 
corresponding  section  of  the  small  vault. 

3.  A  spherical  dome  intersected  by  two  vaults  is  given.     Find  the 
line  of  intersection  both  on  plan  and  elevation. 

4.  A  spherical  dome  and  two  intersecting  vaults.     Find  the  line  of 
intersection  on  plan. 


EXERCISE   SHEET  LXXXVI    (§§194-197) 

Use  large  sheet  or  four  small  ones.     Layout  is  given  on  Fig.  264. 
1.    Find  the  line  of  intersection  of  the  membered  mouldings,  both  on 
plan  and  elevation. 


Fig.  264 

2.  A  vertical  wall  and  a  ceiling  sloping  at  45°  to  H  meet  in  a  corner. 
A  picture  moulding  passes  from  the  vertical  wall  to  the  sloping  ceiling 
and  miters  in  the  corner  on  a  plane  which  bisects  the  angle  between 
the  wall  and  ceiling.  Determine  the  section  for  the  rake  moulding 
and  show  a  full  plan  of  the  mouldings  including  the  line  of  intersection. 

3.  Mouldings  on  a  pediment,  mitering  at  45°.  Determine  the 
section  for  the  raking  moulding. 


302 


DESCRIPTIVE  GEOMETRY 


4.  A  double  raking  moulding,  mitering  on  the  bisecting  plane. 
Determine  the  section  of  the  raking  moulding. 

EXERCISE    SHEET   LXXXVII 

Develop  patterns  for  the  execution  in  sheet  metal  of  the  membered 
mouldings  of  Exercise  LXXXVI,  #1. 

EXERCISE    SHEET   LXXXVIII 

Investigate  the  variation  in  the  right  section  of  the  rake  moulding 
in  Exercise  LXXXVI,  #3,  as  the  angle  of  inclination  of  the  raking  mould- 
ing varies  between  0°  and  90°. 

EXERCISE    SHEET   LXXXIX    (§198) 

Use  either  a  large  or  a  small  sheet.  Vary  scale  to  suit.  Draw  the 
plan,  elevation,  and  two  sections  of  a  spherical  dome  similar  to  Fig. 
221,  but  at  a  larger  scale.  Let  the  dome  be  supported  on  an  hexagonal 
drum  and  spherical  pendentives. 

EXERCISE   SHEET   XC    f§  199) 

Use  two  small  sheets. 

L  Draw  a  half  plan  and  elevation  of  a  spherical  dome,  about  6" 
in  diameter.     On  the  plan  draw  a  pattern  like  that  in  Fig.  222,  making 


E  L  E  V  AT  I  0  H 


Fig.  205 


S ICT I  0  N       Z-Z 


APPENDIX 


303 


the  diagonals  of  the  large  squares  1".     Project  the  pattern  on  the 

elevation. 

2.  After  studying  the  above  problem  with  reference  to  the  dis- 
tortion of  tlie  pattern  on  the  lower  part  of  the  dome,  develop  a  different 
pattern,  based  on  equally  spaced  meridian  sections  and  horizontal 
sections  at  variable  spacings. 


EXERCISE   SHEET   XCI 

1.  Figure  265  shows  a  circular  tower.  The  central  portion  is  covered 
by  a  spherical  dome  and  the  circular  passage  is  covered  by  a  vault  in 
the  form  of  an  annular  torus.  The  door  openings  are  connected  by, 
and  form  a  part  of,  a  conical  vault.  Draw  one  half  of  the  plan,  an 
elevation,  and  a  section  at  j\"  scale,  showing  all  penetration  hues. 

2.  Change  the  conical  vault  in  the  above  problem  to  the  conoidal 
form  shown  in  Fig.  178,  and  repeat. 

EXERCISE   SHEET   XCII 

Figure  266  shows  the  plan  of  a  recessed  doorway.  Let  the  open- 
mgs  A  and  B  each  be  spanned  by  a  semicircular  arch  and  the  space 


Fig.  266 


between  be  covered  by  a  warped  surface.  Draw  a  plan  and  elevation 
at  h"  scale,  indicating  stone  jointing  on  the  basis  of  9  voussoirs  for 
each  arch.  Let  the  maximum  dimension  of  any  stone  be  2'  -  9". 
Show  three  directrices  for  the  warped  surface. 


304  DESCRIPTIVE  GEOMETRY 


i'       6 


EXERCISE  SHEET   XCIII 


Figure  267  shows  the  plan  of  a  semi- 
circular headed  doorw^ay. 

1.   Determine  how  far  the  door  may 
be  opened  before  it  interferes  w4th  the 
^.       Fig.  267  arch  over  the  door.     Scale  1"  =  1'  -  0". 

2.  Determine  a  curve  for  the  arch  soffit  that  vdll  allow  the  door  to 
be  opened  through  90°. 

3.  Determine  an  arrangement  that  will  permit  of  opening  the  door 
through  180°. 

EXERCISE  SHEET   XCIV   (§200) 

Draw  the  projections  of  two  intersecting  barrel  vaults  like  those 
shown  in  Fig.  222,  making  the  larger  48'  in  diameter  and  the  smaller 
40'.  Use  i"  scale.  Draw  on  the  vaults  a  pattern  similar  to  that  in 
the  illustration. 

EXERCISE   SHEET   XCV   (§§202-207) 

1.  A  light  is  situated  at  the  point  I  (77,  21,  19).  Cast  the  shadow 
of  the  line  a  (74,  13,  8),  h  (63,  5,  12). 

2.  Cast  the  shadow  of  the  line  c  (45,  11,  7),  d  (45,  11,  0),  as  cast 
from  the  point  I  in  problem  1  above. 

3.  Cast  the  shadow  of  the  Hne  e  (28,  22,  3),  /  (28,  10,  3),  as  cast  by 
a  light  at  l'  (2,  26,  26). 

4.  Cast  the  shadow  of  the  line  g  (19,  12,  22),  h  (11,  12,  13),  as  cast 
from  the  point  l'  in  problem  3  above. 

EXERCISE   SHEET   XCVI   (§§202-207) 

Cast  the  sun  shadows  of  the  following  Hnes. 

1.  The  line  a  (73,  15,  6),  b  (73,  0,  6).  2.  The  hne  c  (61,  20,  4), 
d  (50,  14,  4).  3.  The  line  e  (43,  14,  11),  /  (43,  14,  0).  4.  The  line 
g  (33,  20,  8),  h  (26,  13,  1).  5.  The  line  k  (20,  2,  20),  I  (6,  12,  3),  and 
another  line  which  passes  through  j  (17,  15,  7)  and  bisects  kl. 

EXERCISE   SHEET   XCVU    (§208) 

1.  The  point  p  (76,  23,  24)  is  a  source  of  light  casting  the  shadow 
of  a  regular  hexagon  with  its  center  at  a  (66,  9,  11),  and  lying  in  a  plane 
parallel  to  H. 


APPENDIX 


305 


2.  Cast  the  sun  shadow  of  a  regular  octahedron,  1|"  on  a  side,  and 
with  ihe  center  of  the  top  face  at  6  (28,  17,  14). 

EXERCISE    SHEET   XCVIII   (§209) 

1.  The  point  a  (53,  27,  27)  is  a  source  of  Hght  and  the  soHd  shown 
in  Fig.  268  is  located  with  the  center  of  its  base  at  b  (58,  0,  7).  Cast 
the  shadow  of  the  solid  on  H  and  V  and  also  on  the  soUd  itself. 


■  o 


^T^ 


1' 


J, 


2i 


8 

i.L 


:4- 


I 
Fig.  268 


5-6 


3*-6" 


6" 
f 


9r- 


FiG.  269 


2.  The  steps  shown  in  Fig.  269  are  located  with  the  point  c  at 
(27,  0,  14).  Cast  the  shadows  as  from  the  source  of  light  given  above. 
Scale,  f"  =  r. 

EXERCISE   SHEET   XCIX    (§209) 
Repeat  Sheet  XCVIII,  casting  the  sun  shadows  of  the  given  objects. 

EXERCISE   SHEET   C    (§210) 

Draw  two  elevations  of  a  dormer  window  similar  to  Fig.  232,  but 
at  a  larger  scale.     Change  the  source  of  light.     Cast  the  shadows. 


EXERCISE    SHEET    CI    (§210) 
Repeat  Sheet  C,  casting  the  sun  shadows. 


306  DESCRIPTIVE  GEOMETRY 

EXERCISE   SHEET   CU   (§211) 

A  circle  2"  in  diameter  has  its  center  at  a  (39,  10,  12),  and  its  plane 
parallel  to  H.  A  source  of  light  is  at  h  (72,  28,  28).  Cast  the  shadow 
of  the  circle. 

EXERCISE   SHEET   CIII   (§211) 

1.  The  point  a  (70,  15,  7)  is  the  center  of  a  circle  2"  in  diameter 
whose  plane  is  parallel  to  V.     Cast  the  sun  shadow. 

2.  The  point  h  (46,  8,  24)  is  the  center  of  a  circle  2"  in  diameter 
whose  plane  is  parallel  to  V.     Cast  the  sun  shadow. 

3.  The  point  c  (26,  8,  8)  is  the  center  of  a  circle  2"  in  diameter 
whose  plane  is  parallel  to  P.     Cast  the  sun  shadow. 

EXERCISE   SHEET   CIV   (§212,1) 

1.  The  point  a  (31,  12,  28)  is  a  source  of  light.  The  point  h  (40, 
17,  15)  is  the  apex  of  a  cone  whose  base  is  a  circle,  parallel  to  H,  and 
2"  in  diameter.  The  center  of  the  base  is  at  e  (50,  4,  10).  Find  the 
shade  and  shadow  of  the  cone. 

2.  The  point  c  (31,  28,  28)  is  a  source  of  hght.  The  point  d  (15, 
16,  8)  is  the  apex  of  a  right  circular  cone  whose  base  is  a  circle  in  H 
and  1^"  in  diameter.     Find  the  shade  and  shadow  of  the  cone. 

EXERCISE   SHEET   CV   (§212,11) 

The  points  a  (66,  5,  14),  h  (35,  5,  14),  and  c  (14,  5,  14)  are  the  base 
centers  for  three  right  circular  cones,  whose  bases  are  each  2"  in  diameter 
and  parallel  to  H.  The  first  is  2"  high,  the  second  is  f "  high,  and  the 
third  is  |"  high.     Cast  the  sun  shadows  of  all  three  cones. 

EXERCISE   SHEET   CVI   (§213) 

The  point  a  (68,  19,  21)  is  a  source  of  light.  The  point  c  (53,  8,  10) 
is  the  center  of  a  sphere  \\"  in  diameter.  Cast  the  shadow  of  the 
sphere. 

EXERCISE   SHEET   CVII   (§213) 

Place  the  sheet  vertically,  HA  in  the  center.  The  point  c  (40,  15, 
16)  is  the  center  of  a  sphere  2"  in  diameter.  Using  the  sun  as  a  source 
of  hght,  determine  the  shade  line  and  cast  the  shadow  of  the  sphere 
on  H  and  V. 

EXERCISE   SHEET   CVIII   (§214) 

Place  the  sheet  vertically,  HA  3^"  from  the  top.  The  point  a  (54, 
24,  25)  is  a  source  of  light;  h  (38,  10,  8)  is  the  center  of  a  sphere  1^" 


APPENDIX 


307 


in  diameter;  c  (50,  11,  16)  d  (36,  20,  12)  is  a  line.  Determine  the 
shade  Une  and  shadow  of  the  sphere,  also  the  shadow  of  cd  on  the 
sphere  and  on  the  planes  of  projection. 

EXERCISE    SHEET    CIX    (§214) 

Cast  the  shadows  on  the  band  stand  shown  in  Fig.  270,  when  the 
source  of  light  is  at  x.     Scale,  Y'  =  1'  —  0". 

EXERCISE    SHEET    CX    (§215) 

Cast  the  shadows  on  the  band  stand  shown  in  Fig.  270,  when  the 
source  of  light  is  at  y.     Scale,  I"  =  V  —  0". 


Fig.  270 


308 


DESCRIPTIVE   GEOMETRY 


EXERCISE    SHEET    CXI    (§216) 

Determine  the  shade  hne  and  shadow,  also  the  shadow  on  //,  for 
the  double-curved  surface  in  Fig.  271.  Use  a  point  source  of  light  on 
the  left  border  line,  11^"  above  H  and  9|"  in  front  of  V. 


Fig.  271 


EXERCISE   SHEET   CXII    (§216) 

With  the  sun  as  the  source  of  light,  determine  the  shade  line  and 
shadows,  also  the  shadows  on  H  and  V,  for  the  double-curved  surface 
in  Fig.  272. 


Fig.  272 


APPENDIX 


309 


EXERCISE   SHEET   CXIII    (§216) 

With  the  sun  as  the  source  of  Hght,  determine 
the  shade  line  and  shadows,  also  the  shadows  on 
H  and  F,  for  the  baluster  in  Fig.  273.  Place 
the  center  of  the  base  on  H,  V  —  6"  from  V 
and  1'  —  6"  from  the  left  border  line.  Scale, 
3"  =  1'  -  0". 

EXERCISE   SHEET   CXIV    (§216) 

Cast  the  sun  shadows  on  a  warped  doorway 
recess,  similar  to  Fig.  223.  In  drawing  the 
doorway,  let  the  front  arch  be  semicircular, 
7'  —  0"  in  diameter;  and  let  the  arch  over  the 
door  be  segmental,  3'  —  0"  radius  and  4'  —  0" 
span.  Let  the  depth  of  the  recess  be  2'  —  6". 
Select  any  convenient  line  for  the  third  directrix 
of  the  warped  surface.  Scale,  I"  =  1'  —  0". 
Small  sheet. 


Fig.  273 


EXERCISE    SHEET    CXV    (§216) 
Cast  the  sun  shadows  on  Fig.  274.     Scale,  \"  =  1'  —  0". 


EXERCISE    SHEET    CXVI    (§220) 


Fig.  274 


Figure  275  gives  the  eleva- 
tions of  various  points  on  a 
certain  piece  of  land  and  the 
course  of  a  stream  crossing  it. 
Draw  the  contour  map  at  a 
scale  of  1"  =  20'  -  0",  indi- 
cating the  contours  at  5'  in- 
tervals. Let  it  be  required  to 
construct  a  road,  20'  wide,  on  a 
level  grade  between  points  A 
and  B.  Assume  the  natural 
slope  of  the  soil  to  be  30°. 
Show  the  cuts  and  embank- 
ments and  the  lines  of  cut  and 

mi. 


310 


DESCRIPTIVE  GEO:^IETIlY 


EXERCISE    SHEET    CXVn    (§220) 

Using  Fig.  275  as  above,  draw  the  contours  and  show  a  road  20' 
wide  between  points  C  and  D.  Let  it  be  built  on  a  12  per  cent  grade. 
Show  the  lines  of  cut  and  fill  and  draw  the  contours  for  the  finished 
embankments. 


EXERCISE    SHEET    CXVIII    (§222) 

Draw  the  isometric  projection  of  the  Celtic 
Cross  in  Fig.  276.  Scale,  f"  =  1'  -  0".  Let  the 
thickness  of  the  shaft  be  1'  —  0". 

EXERCISE    SHEET    CXIX    (§§223,    224) 

Divide  the  sheet  into  three  parts  and  draw  the 
isometric  projections  of  the  following  curved  sur- 
faces :  (1)  A  right  circular  cone,  2¥'  in  diameter 
and  4"  high ;  (2)  a  right  hehcoid,  2"  in  diameter 
and  5"  high;  (3)  a  sphere,  2f"  in  diameter. 


EXERCISE    SHEET    CXX    (§224) 

Draw  the  isometric  projection  of  one  of  the  following  vase  forms : 
Figs.  239,  241,  242,  208,  271,  272,  or  273.  The  envelope  method  will 
be  found  useful. 

EXERCISE    SHEET    CXXI    (§224) 

Draw  the  isometric  projection  of  Fig.  221. 


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